Dgdrxrt

Let $f:Xto Y$ be a morphism of schemes, and let $k(y)$ to be the residue field of the point $y$. The fibre of the morphism $f$ over the point $y$ is defined to be the scheme $X_y=Xtimes Spec(k(y))$.

It's said that $X_y$ is homeomorphic to $f^-1(y)$. If we consider affine schemes, then the point $y$ corresponds to a prime ideal $p$ in $mathcalO_y$, then $f^-1(y)$ should correspond to a prime ideal $q$ in $mathcalO_x$, which is a point. But sometimes $f^-1(y)$ can be more than one point, so I am not sure what $X_y$ looks like.

Besides, it seems that if $H$ is a subscheme of $Y$, then $f^-1(H)$ can be regarded as $Xtimes H$. I hope someone can show me a clear picture. Thanks!

Why should $f^-1(y)$ correspond to only one prime ideal $q$ in $X$?

In the affine case you know that $f : X = mathrmSpec B to mathrmSpec A = Y$ corresponds to a morphism of rings $varphi : A to B$.
Now for $y in Y$ one has $f^-1(y) = x in X mid varphi^-1(x) = y $ which is not necessarily just a single point.

But this is what $f^-1(y)$ looks like in the affine case (I really mean: what the underlying topological space looks like).

One can check that this set corresponds to $mathrmSpec (B otimes_A k(y))$ which has a scheme-structure as well - this is the underlying scheme-structure of $f^-1(y)$ - actually $f^-1(y) = X times_Y mathrmSpec k(y) = mathrmSpec (B otimes_A k(y))$.

Regarding your second question:
One defines $f^-1(H) := X times_Y H$ as the base change of the inclusion $i : H to Y$ along $f$.
Note that the projection $f^-1(H) to X$ is an embedding, as it is the base change of an embedding.

Maybe explicitly writing down the universal property might help you accepting this as a useful and natural definition:

Giving a morphism $Z to f^-1(H)$ is the same as giving morphisms $p : Z to X$ and $q : Z to H$ such that $f circ p = i circ q$.

That is: A morphism $p : Z to X$ factorizes over $f^-1(H)$ if and only if the composition $f circ p : Z to Y$ factorizes over the embedding $H to Y$.

Also note that this actually does the job in for example the categories of sets:

If $f : A to B$ is any map of sets and $B' subseteq B$, then $f^-1(B')$ certainly satisfies the universal property of $A times_B B'$.
If you don't want to use the universal property: $A times_B B' = (a,b') in A times B' mid f(a) = b'$, so that $f^-1(B') = p_1(Atimes_B B')$ where $p_1$ denotes the projection to the first factor.
Also the $b'$ in $(a,b')$ is no real benefit informationwise as by construction $b' = f(a)$, hence one can reconstruct b' just by knowing $a$.

The question is of a local nature so let's assume $X=Spec(B)$ and $Y=Spec (A)$ are affine schemes. Then $f: Spec$ $ Brightarrow Spec$ $A$ corresponds to a ring homomorphism $ g: A rightarrow B$ . Let $y$ correspond to the prime ideal $p$ in $A$ . Then $X_y= X times_ Y Spec (k(y)) = Spec ( B otimes_A frac A_ppA_p) $. Suppose $g^-1(q)=p$ where $qin Spec(B)$ Then $A rightarrow B rightarrow B_q$ factors uniquely through $A_p rightarrow B_q$ and hence we have a unique ring homomorphism from $ B otimes _A A_p/pA_p rightarrow B_q/qB_q$ which is surjective and $B_q/qB_q$ is a field Thus $exists ! zin X_y $ which is the image of the point $Spec (B_q/qB_q)$ .

Now back to geometry.
Consider the commutative diagram

$requireAMScd
beginCD
X_y@>>> Spec (k(y));\
@VVV @VVV \
X@>>> Y;
endCD
$
The maps are all continuous and follow from properties of fibre product. $pi_1(X_y) subset f^-1 (y)$ following commutativity. And the algebra above shows it is a bijection. So you have got a continuous bijection.