Characteristic functions in a CW complex are related by a homeomorphism The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraCharacteristic map and homeomorphism in CW - complexeA covering space of CW complex has an induced CW complex structure.A Confusion Regarding the Intrinsic Definition of CW ComplexesWhy is closure finiteness necessary in this proof that every $n$-cell of an $n$-dimensional CW complex is open?John Lee, CW construction, p139, components of nth skeletonIf $(X, Gamma)$ is a cell complex and $e in Gamma$, then the image of a characteristic map for $e$ equals its closureCW complex definition ambiguityConstructing a regular cell-decomposition of $mathbbS^n$Like a cell complex but characteristic maps send n-cell boundary to other n-cellsAre the restrictions of constructible functions on the level set of smooth map still constructible functions?

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Characteristic functions in a CW complex are related by a homeomorphism



The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraCharacteristic map and homeomorphism in CW - complexeA covering space of CW complex has an induced CW complex structure.A Confusion Regarding the Intrinsic Definition of CW ComplexesWhy is closure finiteness necessary in this proof that every $n$-cell of an $n$-dimensional CW complex is open?John Lee, CW construction, p139, components of nth skeletonIf $(X, Gamma)$ is a cell complex and $e in Gamma$, then the image of a characteristic map for $e$ equals its closureCW complex definition ambiguityConstructing a regular cell-decomposition of $mathbbS^n$Like a cell complex but characteristic maps send n-cell boundary to other n-cellsAre the restrictions of constructible functions on the level set of smooth map still constructible functions?










2












$begingroup$


The book "The Topology of CW Complexes" mentions the following result in passing, but does not provide a proof. Let $(X, mathcalC)$ be a CW complex, $B_k subset mathbbR^k$ be the origin-centered open unit-$k$-ball, and $overlineB_k$ be the corresponding closed $k$-ball. Let $f, g : overlineB_k to X$ both be characteristic functions of the same cell $C in mathcalC$. Then there exists a homeomorphism $h : overlineB_k to overlineB_k$ such that $f = g circ h$. How can you prove this statement?



For a regular cell, the statement is of course trivial, so I am specifically interested in the non-regular case.



EDIT



In the light of the counter-example given by Eric Wofsey, I may have misunderstood what is said in the book. The direct quotes are as follows ($dotE_n$ is the boundary of a closed cell $E_n$).



BEGIN QUOTES



Page 1:



A closed Euclidean $n$-cell $E^n$ is a homeomorphic image of the Euclidean n-cube $I^n$, the cartesian product of $n$ copies of the closed unit interval $I = t in mathbbR : 0 leq t leq 1$.



Page 6:



Let $X$ be a set. A cell structure on $X$ is a pair $(X, Phi)$,
where $Phi$ is a collection of maps of closed Euclidean cells into $X$
satisfying the following conditions.



  1. If $phi in Phi$ and $phi$ has domain $E_n$, then $phi$ is injective on $E_n - dotE_n$.


  2. The images $phi(E_n - dotE_n) : phi in Phi$ partition $X$, i.e., they are disjoint and have union $X$.


  3. If $phi in Phi$ has domain $E_n$, then $phi(dotE_n) subset bigcup psi(E_k - dotE_k) : psi in Phi text has domain E_k text and k leq n - 1$


Page 7:



We say that two cell structures $(X, Phi)$ and $(X, Phi')$ are strictly equivalent if there is a one-to-one correspondence between $Phi$ and $Phi'$ such that a characteristic function with domain $E_n$ corresponds to a characteristic function with domain $E_n$, and corresponding functions differ only by a reparametrization of their domain. That is, if $phi$ and $phi'$ are corresponding functions of $Phi$ and $Phi'$, respectively, then $phi' = phi circ h$, where $h: (E_n, dotE_n) to (E_n, dotE_n)$ is a homeomorphism of pairs. We leave it to the reader to check that this is an equivalence relation on the collection of cell structures on the set X.



If $(X, Phi)$ is a cell structure, let $mathcalS_Phi$ consist of all pairs $(sigma^n, [phi])$, where $sigma^n = phi(E^n)$ and $[phi]$ is the strict equivalence class of $phi in Phi$. For convenience, we will denote such a pair by $sigma^n$ or $phi(E^n)$ with the class of $phi$ understood. Note that if $(X, Phi)$ and $(X, Phi')$ are strictly equivalent cell structures, then $mathcalS_Phi = mathcalS_Phi'$.



Page 8:



A cell complex on a set $X$ or a cellular decomposition
of a set $X$ is an equivalence class of cell structures $(X, Phi)$ under
the equivalence relation of strict equivalence.



A cell complex on $X$ will be denoted by a pair $(X, mathcalS)$, where $mathcalS = S_Phi$ for some representative cell structure $(X, Phi)$. The set $mathcalS$ is called the set of (closed) cells of $(X, mathcalS)$.



Page 41:



A Hausdorff space $X$ is a CW complex with respect to a family of cells $mathcalS$ provided:



  1. the pair $(X,mathcalS)$ is a cell complex such that each cell $sigma in mathcalS$ has a continuous characteristic function;


  2. the space $X$ has the weak topology with respect to $mathcalS$;


  3. the cell complex $(X,mathcalS)$ is closure finite.


END QUOTES



I'm not sure where my misunderstanding is, but I suspect it has to do with that the strict equivalence is defined between cell-structures and not between characteristic functions.



EDIT 2



Referring to the discussions with Paul Frost below, the confusion arises because the definition in the book is more granular than the usual definition of a CW complex. Therefore, if characteristic functions are not related by a homeomorphism in this definition, then they are not the same CW complex.



In contrast, the usual definition of a CW complex requires only the existence of a characteristic function, which makes it possible to provide non-homeomorphically related characteristic functions for the same CW complex, as provided by the counter-example in the accepted answer.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Do you mean Lundell and Weingram's book?
    $endgroup$
    – Paul Frost
    Mar 31 at 8:52










  • $begingroup$
    @PaulFrost Yes.
    $endgroup$
    – kaba
    Mar 31 at 8:54










  • $begingroup$
    @PaulFrost I added more quotes to clarify.
    $endgroup$
    – kaba
    Mar 31 at 10:25










  • $begingroup$
    Could it be that this definition is more granular than the usual definition of a CW complex? If I understand correctly, in this definition if the characteristic functions are not related by a homeomorphism, then they simply aren't the same CW complex. In the usual definition we don't care what the characteristic functions are as long as there is one for each cell. And that would be the origin of my confusion.
    $endgroup$
    – kaba
    Mar 31 at 10:41











  • $begingroup$
    So we find the definition of the set $mathcalS$ of (closed) cells on p.8. Although it is not explicitly mentioned, a cell will be a member of $mathcalS$. Thus, a cell consists of a subspace $sigma^n subset X$ plus an equivalence class of characteristic maps. This is somewhat unusual, most authors only require the existence of a characteristic map. And in fact, the book says on p.41 "such that each cell $Sigma in mathcalS$ has a continuous characteristic function". This would be redundant if we adopt the definition on p.8. I guess the book is not completely consistent.
    $endgroup$
    – Paul Frost
    Mar 31 at 10:43















2












$begingroup$


The book "The Topology of CW Complexes" mentions the following result in passing, but does not provide a proof. Let $(X, mathcalC)$ be a CW complex, $B_k subset mathbbR^k$ be the origin-centered open unit-$k$-ball, and $overlineB_k$ be the corresponding closed $k$-ball. Let $f, g : overlineB_k to X$ both be characteristic functions of the same cell $C in mathcalC$. Then there exists a homeomorphism $h : overlineB_k to overlineB_k$ such that $f = g circ h$. How can you prove this statement?



For a regular cell, the statement is of course trivial, so I am specifically interested in the non-regular case.



EDIT



In the light of the counter-example given by Eric Wofsey, I may have misunderstood what is said in the book. The direct quotes are as follows ($dotE_n$ is the boundary of a closed cell $E_n$).



BEGIN QUOTES



Page 1:



A closed Euclidean $n$-cell $E^n$ is a homeomorphic image of the Euclidean n-cube $I^n$, the cartesian product of $n$ copies of the closed unit interval $I = t in mathbbR : 0 leq t leq 1$.



Page 6:



Let $X$ be a set. A cell structure on $X$ is a pair $(X, Phi)$,
where $Phi$ is a collection of maps of closed Euclidean cells into $X$
satisfying the following conditions.



  1. If $phi in Phi$ and $phi$ has domain $E_n$, then $phi$ is injective on $E_n - dotE_n$.


  2. The images $phi(E_n - dotE_n) : phi in Phi$ partition $X$, i.e., they are disjoint and have union $X$.


  3. If $phi in Phi$ has domain $E_n$, then $phi(dotE_n) subset bigcup psi(E_k - dotE_k) : psi in Phi text has domain E_k text and k leq n - 1$


Page 7:



We say that two cell structures $(X, Phi)$ and $(X, Phi')$ are strictly equivalent if there is a one-to-one correspondence between $Phi$ and $Phi'$ such that a characteristic function with domain $E_n$ corresponds to a characteristic function with domain $E_n$, and corresponding functions differ only by a reparametrization of their domain. That is, if $phi$ and $phi'$ are corresponding functions of $Phi$ and $Phi'$, respectively, then $phi' = phi circ h$, where $h: (E_n, dotE_n) to (E_n, dotE_n)$ is a homeomorphism of pairs. We leave it to the reader to check that this is an equivalence relation on the collection of cell structures on the set X.



If $(X, Phi)$ is a cell structure, let $mathcalS_Phi$ consist of all pairs $(sigma^n, [phi])$, where $sigma^n = phi(E^n)$ and $[phi]$ is the strict equivalence class of $phi in Phi$. For convenience, we will denote such a pair by $sigma^n$ or $phi(E^n)$ with the class of $phi$ understood. Note that if $(X, Phi)$ and $(X, Phi')$ are strictly equivalent cell structures, then $mathcalS_Phi = mathcalS_Phi'$.



Page 8:



A cell complex on a set $X$ or a cellular decomposition
of a set $X$ is an equivalence class of cell structures $(X, Phi)$ under
the equivalence relation of strict equivalence.



A cell complex on $X$ will be denoted by a pair $(X, mathcalS)$, where $mathcalS = S_Phi$ for some representative cell structure $(X, Phi)$. The set $mathcalS$ is called the set of (closed) cells of $(X, mathcalS)$.



Page 41:



A Hausdorff space $X$ is a CW complex with respect to a family of cells $mathcalS$ provided:



  1. the pair $(X,mathcalS)$ is a cell complex such that each cell $sigma in mathcalS$ has a continuous characteristic function;


  2. the space $X$ has the weak topology with respect to $mathcalS$;


  3. the cell complex $(X,mathcalS)$ is closure finite.


END QUOTES



I'm not sure where my misunderstanding is, but I suspect it has to do with that the strict equivalence is defined between cell-structures and not between characteristic functions.



EDIT 2



Referring to the discussions with Paul Frost below, the confusion arises because the definition in the book is more granular than the usual definition of a CW complex. Therefore, if characteristic functions are not related by a homeomorphism in this definition, then they are not the same CW complex.



In contrast, the usual definition of a CW complex requires only the existence of a characteristic function, which makes it possible to provide non-homeomorphically related characteristic functions for the same CW complex, as provided by the counter-example in the accepted answer.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Do you mean Lundell and Weingram's book?
    $endgroup$
    – Paul Frost
    Mar 31 at 8:52










  • $begingroup$
    @PaulFrost Yes.
    $endgroup$
    – kaba
    Mar 31 at 8:54










  • $begingroup$
    @PaulFrost I added more quotes to clarify.
    $endgroup$
    – kaba
    Mar 31 at 10:25










  • $begingroup$
    Could it be that this definition is more granular than the usual definition of a CW complex? If I understand correctly, in this definition if the characteristic functions are not related by a homeomorphism, then they simply aren't the same CW complex. In the usual definition we don't care what the characteristic functions are as long as there is one for each cell. And that would be the origin of my confusion.
    $endgroup$
    – kaba
    Mar 31 at 10:41











  • $begingroup$
    So we find the definition of the set $mathcalS$ of (closed) cells on p.8. Although it is not explicitly mentioned, a cell will be a member of $mathcalS$. Thus, a cell consists of a subspace $sigma^n subset X$ plus an equivalence class of characteristic maps. This is somewhat unusual, most authors only require the existence of a characteristic map. And in fact, the book says on p.41 "such that each cell $Sigma in mathcalS$ has a continuous characteristic function". This would be redundant if we adopt the definition on p.8. I guess the book is not completely consistent.
    $endgroup$
    – Paul Frost
    Mar 31 at 10:43













2












2








2





$begingroup$


The book "The Topology of CW Complexes" mentions the following result in passing, but does not provide a proof. Let $(X, mathcalC)$ be a CW complex, $B_k subset mathbbR^k$ be the origin-centered open unit-$k$-ball, and $overlineB_k$ be the corresponding closed $k$-ball. Let $f, g : overlineB_k to X$ both be characteristic functions of the same cell $C in mathcalC$. Then there exists a homeomorphism $h : overlineB_k to overlineB_k$ such that $f = g circ h$. How can you prove this statement?



For a regular cell, the statement is of course trivial, so I am specifically interested in the non-regular case.



EDIT



In the light of the counter-example given by Eric Wofsey, I may have misunderstood what is said in the book. The direct quotes are as follows ($dotE_n$ is the boundary of a closed cell $E_n$).



BEGIN QUOTES



Page 1:



A closed Euclidean $n$-cell $E^n$ is a homeomorphic image of the Euclidean n-cube $I^n$, the cartesian product of $n$ copies of the closed unit interval $I = t in mathbbR : 0 leq t leq 1$.



Page 6:



Let $X$ be a set. A cell structure on $X$ is a pair $(X, Phi)$,
where $Phi$ is a collection of maps of closed Euclidean cells into $X$
satisfying the following conditions.



  1. If $phi in Phi$ and $phi$ has domain $E_n$, then $phi$ is injective on $E_n - dotE_n$.


  2. The images $phi(E_n - dotE_n) : phi in Phi$ partition $X$, i.e., they are disjoint and have union $X$.


  3. If $phi in Phi$ has domain $E_n$, then $phi(dotE_n) subset bigcup psi(E_k - dotE_k) : psi in Phi text has domain E_k text and k leq n - 1$


Page 7:



We say that two cell structures $(X, Phi)$ and $(X, Phi')$ are strictly equivalent if there is a one-to-one correspondence between $Phi$ and $Phi'$ such that a characteristic function with domain $E_n$ corresponds to a characteristic function with domain $E_n$, and corresponding functions differ only by a reparametrization of their domain. That is, if $phi$ and $phi'$ are corresponding functions of $Phi$ and $Phi'$, respectively, then $phi' = phi circ h$, where $h: (E_n, dotE_n) to (E_n, dotE_n)$ is a homeomorphism of pairs. We leave it to the reader to check that this is an equivalence relation on the collection of cell structures on the set X.



If $(X, Phi)$ is a cell structure, let $mathcalS_Phi$ consist of all pairs $(sigma^n, [phi])$, where $sigma^n = phi(E^n)$ and $[phi]$ is the strict equivalence class of $phi in Phi$. For convenience, we will denote such a pair by $sigma^n$ or $phi(E^n)$ with the class of $phi$ understood. Note that if $(X, Phi)$ and $(X, Phi')$ are strictly equivalent cell structures, then $mathcalS_Phi = mathcalS_Phi'$.



Page 8:



A cell complex on a set $X$ or a cellular decomposition
of a set $X$ is an equivalence class of cell structures $(X, Phi)$ under
the equivalence relation of strict equivalence.



A cell complex on $X$ will be denoted by a pair $(X, mathcalS)$, where $mathcalS = S_Phi$ for some representative cell structure $(X, Phi)$. The set $mathcalS$ is called the set of (closed) cells of $(X, mathcalS)$.



Page 41:



A Hausdorff space $X$ is a CW complex with respect to a family of cells $mathcalS$ provided:



  1. the pair $(X,mathcalS)$ is a cell complex such that each cell $sigma in mathcalS$ has a continuous characteristic function;


  2. the space $X$ has the weak topology with respect to $mathcalS$;


  3. the cell complex $(X,mathcalS)$ is closure finite.


END QUOTES



I'm not sure where my misunderstanding is, but I suspect it has to do with that the strict equivalence is defined between cell-structures and not between characteristic functions.



EDIT 2



Referring to the discussions with Paul Frost below, the confusion arises because the definition in the book is more granular than the usual definition of a CW complex. Therefore, if characteristic functions are not related by a homeomorphism in this definition, then they are not the same CW complex.



In contrast, the usual definition of a CW complex requires only the existence of a characteristic function, which makes it possible to provide non-homeomorphically related characteristic functions for the same CW complex, as provided by the counter-example in the accepted answer.










share|cite|improve this question











$endgroup$




The book "The Topology of CW Complexes" mentions the following result in passing, but does not provide a proof. Let $(X, mathcalC)$ be a CW complex, $B_k subset mathbbR^k$ be the origin-centered open unit-$k$-ball, and $overlineB_k$ be the corresponding closed $k$-ball. Let $f, g : overlineB_k to X$ both be characteristic functions of the same cell $C in mathcalC$. Then there exists a homeomorphism $h : overlineB_k to overlineB_k$ such that $f = g circ h$. How can you prove this statement?



For a regular cell, the statement is of course trivial, so I am specifically interested in the non-regular case.



EDIT



In the light of the counter-example given by Eric Wofsey, I may have misunderstood what is said in the book. The direct quotes are as follows ($dotE_n$ is the boundary of a closed cell $E_n$).



BEGIN QUOTES



Page 1:



A closed Euclidean $n$-cell $E^n$ is a homeomorphic image of the Euclidean n-cube $I^n$, the cartesian product of $n$ copies of the closed unit interval $I = t in mathbbR : 0 leq t leq 1$.



Page 6:



Let $X$ be a set. A cell structure on $X$ is a pair $(X, Phi)$,
where $Phi$ is a collection of maps of closed Euclidean cells into $X$
satisfying the following conditions.



  1. If $phi in Phi$ and $phi$ has domain $E_n$, then $phi$ is injective on $E_n - dotE_n$.


  2. The images $phi(E_n - dotE_n) : phi in Phi$ partition $X$, i.e., they are disjoint and have union $X$.


  3. If $phi in Phi$ has domain $E_n$, then $phi(dotE_n) subset bigcup psi(E_k - dotE_k) : psi in Phi text has domain E_k text and k leq n - 1$


Page 7:



We say that two cell structures $(X, Phi)$ and $(X, Phi')$ are strictly equivalent if there is a one-to-one correspondence between $Phi$ and $Phi'$ such that a characteristic function with domain $E_n$ corresponds to a characteristic function with domain $E_n$, and corresponding functions differ only by a reparametrization of their domain. That is, if $phi$ and $phi'$ are corresponding functions of $Phi$ and $Phi'$, respectively, then $phi' = phi circ h$, where $h: (E_n, dotE_n) to (E_n, dotE_n)$ is a homeomorphism of pairs. We leave it to the reader to check that this is an equivalence relation on the collection of cell structures on the set X.



If $(X, Phi)$ is a cell structure, let $mathcalS_Phi$ consist of all pairs $(sigma^n, [phi])$, where $sigma^n = phi(E^n)$ and $[phi]$ is the strict equivalence class of $phi in Phi$. For convenience, we will denote such a pair by $sigma^n$ or $phi(E^n)$ with the class of $phi$ understood. Note that if $(X, Phi)$ and $(X, Phi')$ are strictly equivalent cell structures, then $mathcalS_Phi = mathcalS_Phi'$.



Page 8:



A cell complex on a set $X$ or a cellular decomposition
of a set $X$ is an equivalence class of cell structures $(X, Phi)$ under
the equivalence relation of strict equivalence.



A cell complex on $X$ will be denoted by a pair $(X, mathcalS)$, where $mathcalS = S_Phi$ for some representative cell structure $(X, Phi)$. The set $mathcalS$ is called the set of (closed) cells of $(X, mathcalS)$.



Page 41:



A Hausdorff space $X$ is a CW complex with respect to a family of cells $mathcalS$ provided:



  1. the pair $(X,mathcalS)$ is a cell complex such that each cell $sigma in mathcalS$ has a continuous characteristic function;


  2. the space $X$ has the weak topology with respect to $mathcalS$;


  3. the cell complex $(X,mathcalS)$ is closure finite.


END QUOTES



I'm not sure where my misunderstanding is, but I suspect it has to do with that the strict equivalence is defined between cell-structures and not between characteristic functions.



EDIT 2



Referring to the discussions with Paul Frost below, the confusion arises because the definition in the book is more granular than the usual definition of a CW complex. Therefore, if characteristic functions are not related by a homeomorphism in this definition, then they are not the same CW complex.



In contrast, the usual definition of a CW complex requires only the existence of a characteristic function, which makes it possible to provide non-homeomorphically related characteristic functions for the same CW complex, as provided by the counter-example in the accepted answer.







general-topology algebraic-topology cw-complexes






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 31 at 10:55







kaba

















asked Mar 31 at 8:08









kabakaba

27417




27417











  • $begingroup$
    Do you mean Lundell and Weingram's book?
    $endgroup$
    – Paul Frost
    Mar 31 at 8:52










  • $begingroup$
    @PaulFrost Yes.
    $endgroup$
    – kaba
    Mar 31 at 8:54










  • $begingroup$
    @PaulFrost I added more quotes to clarify.
    $endgroup$
    – kaba
    Mar 31 at 10:25










  • $begingroup$
    Could it be that this definition is more granular than the usual definition of a CW complex? If I understand correctly, in this definition if the characteristic functions are not related by a homeomorphism, then they simply aren't the same CW complex. In the usual definition we don't care what the characteristic functions are as long as there is one for each cell. And that would be the origin of my confusion.
    $endgroup$
    – kaba
    Mar 31 at 10:41











  • $begingroup$
    So we find the definition of the set $mathcalS$ of (closed) cells on p.8. Although it is not explicitly mentioned, a cell will be a member of $mathcalS$. Thus, a cell consists of a subspace $sigma^n subset X$ plus an equivalence class of characteristic maps. This is somewhat unusual, most authors only require the existence of a characteristic map. And in fact, the book says on p.41 "such that each cell $Sigma in mathcalS$ has a continuous characteristic function". This would be redundant if we adopt the definition on p.8. I guess the book is not completely consistent.
    $endgroup$
    – Paul Frost
    Mar 31 at 10:43
















  • $begingroup$
    Do you mean Lundell and Weingram's book?
    $endgroup$
    – Paul Frost
    Mar 31 at 8:52










  • $begingroup$
    @PaulFrost Yes.
    $endgroup$
    – kaba
    Mar 31 at 8:54










  • $begingroup$
    @PaulFrost I added more quotes to clarify.
    $endgroup$
    – kaba
    Mar 31 at 10:25










  • $begingroup$
    Could it be that this definition is more granular than the usual definition of a CW complex? If I understand correctly, in this definition if the characteristic functions are not related by a homeomorphism, then they simply aren't the same CW complex. In the usual definition we don't care what the characteristic functions are as long as there is one for each cell. And that would be the origin of my confusion.
    $endgroup$
    – kaba
    Mar 31 at 10:41











  • $begingroup$
    So we find the definition of the set $mathcalS$ of (closed) cells on p.8. Although it is not explicitly mentioned, a cell will be a member of $mathcalS$. Thus, a cell consists of a subspace $sigma^n subset X$ plus an equivalence class of characteristic maps. This is somewhat unusual, most authors only require the existence of a characteristic map. And in fact, the book says on p.41 "such that each cell $Sigma in mathcalS$ has a continuous characteristic function". This would be redundant if we adopt the definition on p.8. I guess the book is not completely consistent.
    $endgroup$
    – Paul Frost
    Mar 31 at 10:43















$begingroup$
Do you mean Lundell and Weingram's book?
$endgroup$
– Paul Frost
Mar 31 at 8:52




$begingroup$
Do you mean Lundell and Weingram's book?
$endgroup$
– Paul Frost
Mar 31 at 8:52












$begingroup$
@PaulFrost Yes.
$endgroup$
– kaba
Mar 31 at 8:54




$begingroup$
@PaulFrost Yes.
$endgroup$
– kaba
Mar 31 at 8:54












$begingroup$
@PaulFrost I added more quotes to clarify.
$endgroup$
– kaba
Mar 31 at 10:25




$begingroup$
@PaulFrost I added more quotes to clarify.
$endgroup$
– kaba
Mar 31 at 10:25












$begingroup$
Could it be that this definition is more granular than the usual definition of a CW complex? If I understand correctly, in this definition if the characteristic functions are not related by a homeomorphism, then they simply aren't the same CW complex. In the usual definition we don't care what the characteristic functions are as long as there is one for each cell. And that would be the origin of my confusion.
$endgroup$
– kaba
Mar 31 at 10:41





$begingroup$
Could it be that this definition is more granular than the usual definition of a CW complex? If I understand correctly, in this definition if the characteristic functions are not related by a homeomorphism, then they simply aren't the same CW complex. In the usual definition we don't care what the characteristic functions are as long as there is one for each cell. And that would be the origin of my confusion.
$endgroup$
– kaba
Mar 31 at 10:41













$begingroup$
So we find the definition of the set $mathcalS$ of (closed) cells on p.8. Although it is not explicitly mentioned, a cell will be a member of $mathcalS$. Thus, a cell consists of a subspace $sigma^n subset X$ plus an equivalence class of characteristic maps. This is somewhat unusual, most authors only require the existence of a characteristic map. And in fact, the book says on p.41 "such that each cell $Sigma in mathcalS$ has a continuous characteristic function". This would be redundant if we adopt the definition on p.8. I guess the book is not completely consistent.
$endgroup$
– Paul Frost
Mar 31 at 10:43




$begingroup$
So we find the definition of the set $mathcalS$ of (closed) cells on p.8. Although it is not explicitly mentioned, a cell will be a member of $mathcalS$. Thus, a cell consists of a subspace $sigma^n subset X$ plus an equivalence class of characteristic maps. This is somewhat unusual, most authors only require the existence of a characteristic map. And in fact, the book says on p.41 "such that each cell $Sigma in mathcalS$ has a continuous characteristic function". This would be redundant if we adopt the definition on p.8. I guess the book is not completely consistent.
$endgroup$
– Paul Frost
Mar 31 at 10:43










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This is false. If $k>1$, then there exists a map $f:overlineB_ktooverlineB_k$ which restricts to a homeomorphism $B_kto B_k$ and maps $partial B_k$ to itself but which is not injective on $partial B_k$. So, taking $X=overlineB_k$ with a CW-complex structure that has $B_k$ as a $k$-cell, both $f$ and $g=1_overlineB_k$ are possible characteristic functions for the $k$-cell. Since $f$ is not a homeomorphism, there cannot exist any homeomorphism $h:overlineB_ktooverlineB_k$ such that $f=gcirc h$.



For an explicit example of such an $f$ for $k=2$, let $S=[0,1]times[0,1]$ and $T=(x,y)inmathbbR^2:0leq yleq xleq 1$ and define $f_0:Sto T$ by $f_0(x,y)=(x,xy)$. Then $f_0$ is a homeomorphism on the interiors of $S$ and $T$ but collapses one of the sides of $S$ to a single point. Composing with homeomorphisms $S,Tcong overlineB_2$ we get $f:overlineB_2tooverlineB_2$ which is a homeomorphism on the interior but collapses an arc on the boundary to a point.






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    $begingroup$

    This is false. If $k>1$, then there exists a map $f:overlineB_ktooverlineB_k$ which restricts to a homeomorphism $B_kto B_k$ and maps $partial B_k$ to itself but which is not injective on $partial B_k$. So, taking $X=overlineB_k$ with a CW-complex structure that has $B_k$ as a $k$-cell, both $f$ and $g=1_overlineB_k$ are possible characteristic functions for the $k$-cell. Since $f$ is not a homeomorphism, there cannot exist any homeomorphism $h:overlineB_ktooverlineB_k$ such that $f=gcirc h$.



    For an explicit example of such an $f$ for $k=2$, let $S=[0,1]times[0,1]$ and $T=(x,y)inmathbbR^2:0leq yleq xleq 1$ and define $f_0:Sto T$ by $f_0(x,y)=(x,xy)$. Then $f_0$ is a homeomorphism on the interiors of $S$ and $T$ but collapses one of the sides of $S$ to a single point. Composing with homeomorphisms $S,Tcong overlineB_2$ we get $f:overlineB_2tooverlineB_2$ which is a homeomorphism on the interior but collapses an arc on the boundary to a point.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      This is false. If $k>1$, then there exists a map $f:overlineB_ktooverlineB_k$ which restricts to a homeomorphism $B_kto B_k$ and maps $partial B_k$ to itself but which is not injective on $partial B_k$. So, taking $X=overlineB_k$ with a CW-complex structure that has $B_k$ as a $k$-cell, both $f$ and $g=1_overlineB_k$ are possible characteristic functions for the $k$-cell. Since $f$ is not a homeomorphism, there cannot exist any homeomorphism $h:overlineB_ktooverlineB_k$ such that $f=gcirc h$.



      For an explicit example of such an $f$ for $k=2$, let $S=[0,1]times[0,1]$ and $T=(x,y)inmathbbR^2:0leq yleq xleq 1$ and define $f_0:Sto T$ by $f_0(x,y)=(x,xy)$. Then $f_0$ is a homeomorphism on the interiors of $S$ and $T$ but collapses one of the sides of $S$ to a single point. Composing with homeomorphisms $S,Tcong overlineB_2$ we get $f:overlineB_2tooverlineB_2$ which is a homeomorphism on the interior but collapses an arc on the boundary to a point.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        This is false. If $k>1$, then there exists a map $f:overlineB_ktooverlineB_k$ which restricts to a homeomorphism $B_kto B_k$ and maps $partial B_k$ to itself but which is not injective on $partial B_k$. So, taking $X=overlineB_k$ with a CW-complex structure that has $B_k$ as a $k$-cell, both $f$ and $g=1_overlineB_k$ are possible characteristic functions for the $k$-cell. Since $f$ is not a homeomorphism, there cannot exist any homeomorphism $h:overlineB_ktooverlineB_k$ such that $f=gcirc h$.



        For an explicit example of such an $f$ for $k=2$, let $S=[0,1]times[0,1]$ and $T=(x,y)inmathbbR^2:0leq yleq xleq 1$ and define $f_0:Sto T$ by $f_0(x,y)=(x,xy)$. Then $f_0$ is a homeomorphism on the interiors of $S$ and $T$ but collapses one of the sides of $S$ to a single point. Composing with homeomorphisms $S,Tcong overlineB_2$ we get $f:overlineB_2tooverlineB_2$ which is a homeomorphism on the interior but collapses an arc on the boundary to a point.






        share|cite|improve this answer









        $endgroup$



        This is false. If $k>1$, then there exists a map $f:overlineB_ktooverlineB_k$ which restricts to a homeomorphism $B_kto B_k$ and maps $partial B_k$ to itself but which is not injective on $partial B_k$. So, taking $X=overlineB_k$ with a CW-complex structure that has $B_k$ as a $k$-cell, both $f$ and $g=1_overlineB_k$ are possible characteristic functions for the $k$-cell. Since $f$ is not a homeomorphism, there cannot exist any homeomorphism $h:overlineB_ktooverlineB_k$ such that $f=gcirc h$.



        For an explicit example of such an $f$ for $k=2$, let $S=[0,1]times[0,1]$ and $T=(x,y)inmathbbR^2:0leq yleq xleq 1$ and define $f_0:Sto T$ by $f_0(x,y)=(x,xy)$. Then $f_0$ is a homeomorphism on the interiors of $S$ and $T$ but collapses one of the sides of $S$ to a single point. Composing with homeomorphisms $S,Tcong overlineB_2$ we get $f:overlineB_2tooverlineB_2$ which is a homeomorphism on the interior but collapses an arc on the boundary to a point.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 31 at 8:48









        Eric WofseyEric Wofsey

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