Let $f: X \to Y$ be a morphism of schemes, take $y \in Y$ and let $k$ be the residue field of $y$. We also have $i_y: \operatorname{Spec}{k} \to Y$. Then we can form a fiber product $Z$ which is the fiber product of $f$ over $y$.
I would like to know how to deduce the following facts.
1) How do we know that as a topological space $Z$ is $f^{-1}(y)$?
2) Take any $t \in f^{-1}(y)$. Is it always the case that $O_{Z, t} = O_{X, t}$?
Thank you!
As a first step, reduce to the affine case. To start, we may assume $Y$ is affine: factoring $i_y$ as the composition of $\operatorname{Spec} k \to \operatorname{Spec} B\to Y$ we see that the fiber product of $i_y$ and $f$ is the same as the fiber product of $i_y':\operatorname{Spec} k\to \operatorname{Spec} B$ and $\overline{f}:X_{\operatorname{Spec} B} \to \operatorname{Spec} B$. Next, we can cover $X$ by open affines with open immersions $i_A:\operatorname{Spec} A\to Y$. As open immersions are stable under base change, we get that the morphism $\operatorname{Spec} A\times_Y \{y\}\to X_y$ is also an open immersion, and the schemes $\operatorname{Spec} A\times_Y y$ cover $X_y$. So it's enough to understand the affine situation.
In the affine case, our map $X\to Y$ can be represented as a ring map $f: B\to A$ where a point $\mathfrak{p}\in X$ maps to a point $f^{-1}(\mathfrak{p})\in Y$. Let $\mathfrak{q}$ be the prime ideal associated to the point $y$. We will proceed to compute the fiber product by intermediate steps (recalling that for a fiber diagram of affine schemes $\operatorname{Spec} R\to \operatorname{Spec} S$ and $\operatorname{Spec} T\to \operatorname{Spec} S$, the fiber product is the scheme $\operatorname{Spec} R\otimes_S T$ with the obvious natural maps). Consider the following diagram:
$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccc} Z & \ra{} & X_{\operatorname{Spec} B_\mathfrak{q}} & \ra{} & X=\operatorname{Spec} A \\ \da{} & & \da{} & & \da{} \\ \{y\} & \ra{} & \operatorname{Spec} B_\mathfrak{q} & \ra{} & Y=\operatorname{Spec} B\\ \end{array} $$
Our first step is to localize both $A$ and $B$ at the ideal $\mathfrak{q}\subset B$. Geometrically, this corresponds to taking the fiber product along the map $\operatorname{Spec} B_\mathfrak{q} \to \operatorname{Spec} B$ in the above diagram. We then identify $X_{\operatorname{Spec} B_\mathfrak{q}} = \operatorname{Spec} A \otimes_B B_\mathfrak{q} = \operatorname{Spec} A_\mathfrak{q}$. We note that by the description of the ideals of $A_\mathfrak{q}$, this preserves the points in the fiber over $y$: prime ideals of the localization $A_\mathfrak{q}$ are exactly those that do not intersect the set we're localizing at, which is equivalent to the preimages of those prime ideals being disjoint from the complement of $\mathfrak{q}$, or equivalently contained in $\mathfrak{q}$.
We are now in the situation of $B_\mathfrak{q} \to A\otimes_B B_\mathfrak{q} := A_\mathfrak{q}$. In order to compute the next fiber product, we recognize the inclusion $\{y\} \to \operatorname{Spec} B_\mathfrak{q}$ as the map $\operatorname{Spec} \kappa(y) = \operatorname{Spec} B_\mathfrak{q}/\mathfrak{q}B_\mathfrak{q} \to \operatorname{Spec} B_\mathfrak{q}$. Therefore $X_y=\operatorname{Spec} A_\mathfrak{q}/\mathfrak{q}A_\mathfrak{q}$, and it remains to identify the prime ideals of this ring. By the correspondence theorem, these are exactly the prime ideals that contain $\mathfrak{q}A_\mathfrak{q}$. But these are precisely the prime ideals who's preimage under the map $B_\mathfrak{q}\to A_\mathfrak{q}$ is exactly $\mathfrak{q}$: that is, the points of $\operatorname{Spec} A_\mathfrak{q}$ which are mapped to $\{y\}$. So this step also preserves the points in the fiber over $y$.
As these two steps preserve the points in the fiber over $y$ and combine to produce the fiber over $y$, we are done: the scheme-theoretic fiber over $y$ has the same underlying topological space as the topological fiber over $y$.
For an alternate proof, points of the fiber product $X\times_S Y$ correspond to quadruples $(x,y,s,\mathfrak{p})$ with $f(x)=g(y)=s$ and $\mathfrak{p}$ a prime ideal of $\operatorname{Spec} \kappa(x)\otimes_{\kappa(s)} \kappa(y)$. In our case, $\kappa(s)=\kappa(y)$ and so the tensor product is really just $\kappa(x)$, the prime ideal must be zero, and we see the correspondence we're after.
For question 2, I fear you did not actually do an example - I suggested a linear projection between affine spaces in the comments, and I stand by that assertion! Please try it yourself before looking at the details contained under the following spoiler:
This is false in general. Consider $\Bbb A^2_k \to \Bbb A^1_k$ by the natural projection map. Let $t$ and $y$ both be the origins in their respective spaces. Then $\mathcal{O}_{X,t}\cong k[u,v]_{(u,v)}$, while $\mathcal{O}_{Z,t} \cong k[v]_{(v)}$, which are obviously not isomorphic - their maximal ideals have different numbers of generators, for one. In general, this is rare - it requires $y$ being an isolated point or $t$ having an open neighborhood inside $X$ which is entirely contained inside $Z$.
