Pullback of a function along identity? If $f:A\rightarrow B$ is a function, then $A\times _BB= \left\{ (a,b)\mid f(a)=b\right\}$. Isn't this the preimage of the image of $f$? I.e is $A\times_BB\cong f^\ast f_\ast(A)$?
If working with spaces, is this a homeomorphism?
 A: Let $C$ be a category and $f\colon a \to b$ and $g\colon c \to b$ two arrows of $C$. Suppose the pullback square
$$
\require{AMScd}
\begin{CD}
p @>{f'}>> c\\
@VV{g'}V @VV{g}V \\
a @>{f}>> b;
\end{CD}
$$
is representable in $C$, i.e. the square is cartesian. It is straightforward to show that if $g$ is an isomorphism, then also $g'$ is so (see, for instance, Proposition 2.5.3 (ii) of F. Borceux, Handbook of Categorical Algebra I).
In the case where $g$ is the identity of $b$, we can choose $f' = f$ and $g' = 1_a$, i.e. the following square
$$
\require{AMScd}
\begin{CD}
a @>{f}>> b\\
@VV{1_a}V @VV{1_b}V \\
a @>{f}>> b;
\end{CD}
$$
is cartesian. One can actually deduce this easily from the fact stated above. Let's see a direct proof anyway.
Suppose we have a pair of arrows $h\colon q \to a$ and $h' \colon q \to b$ such that $h' = f\circ h$. Thus $h \colon q \to a$ is an arrow such that $1_ah$ and $fh =h'$. We must show it is the unique with this property. Indeed, the only arrow $k\colon q \to a$ which satisfies $1_a k = h$ is $h$ and we are done.
In your particular case, with the standard construction of pullbacks in $\mathbf{Set}$ you get precisely the graph $\{(a, b) \in A\times B : f(a)= b\}$ of the function $f$ as the set $A\times_B B$, which is canonically isomorphic to $A$ itself.
