*This question was originally posted last week in Math Stack Exchange (see [here](https://math.stackexchange.com/questions/4483240/compatibility-of-pullbacks-with-an-equivalence-relation)).*


I'm currently working on the proof of the existence of the sheafification in Angelo Vistoli’s 2007 [*Notes on Grothendieck topologies,
fibered categories
and descent theory*](https://arxiv.org/pdf/math/0412512.pdf), but i currently stuck on a statement in the proof of Theorem 2.64. $(ii)$. That is the following:


Note: In this context, a Grothendieck topology is a Singleton Grothendieck topology, such that every covering of an object $U$ in $C$ is a single map $\phi:T\rightarrow U$.


We have a site $C$, that is a category $C$ with a Grothendieck topology. Then we have a functor $$F:C^{op}\rightarrow Set$$
We define an equivalence relation $\sim$ for every object $U$ of $C$ on $F(U)$ as: $a\sim b$ if there exists a covering $\phi:T\rightarrow U$, such that the pullback $F(\phi)=\phi^\*$ of $a$ and $b$ coincide in $F(T)$. In other words $\phi^\*(a)=\phi^\*(b)$.  

Now the statement is, that for every morphism $f:S\rightarrow U$ the pullback $F(f)=f^\*:F(U)\rightarrow F(S)$ is compatible with $\sim$. That means:
$$
a,b\in F(U):a\sim b\Rightarrow f^\*(a)\sim f^\*(b)
$$


**Question:** How do I prove this?



Compatibility of pullbacks with an equivalence relation
After I saw my mistake I got the proof really quick. We need the fibre product of $f$ and $\phi$, where $f$ is an arbitrarily morphism and $\phi$ the covering from $a\sim b$.
So we get the following commutative diagram:
$\require{AMScd}$
\begin{CD}
S\times\_UT @>{pr\_2}>> T\\
@V{pr\_1}VV @VV{\phi}V\\
S @>{f}>> U
\end{CD}


hence
$$
\phi\circ pr\_2=f\circ pr\_1.
$$
Using $F$ on both sides we get
$$
F(\phi\circ pr\_2)=F(f\circ pr\_1)\quad\text{ or }\quad pr\_2^\*\circ\phi^\*=pr\_1^\*\circ f^\*.
$$
Hence, $pr\_1^\*(f^\*(a))=(pr\_1^\*\circ f^\*)(a)=pr\_2^\*(\phi^\*(a))=pr\_2^\*(\phi^\*(b))=(pr\_1^\*\circ f^\*)(b)=pr\_1^\*(f^\*(b)).$


Since $pr\_1$ is a covering (Covering Axioms), we get $f^\*(a)\sim f^\*(b)$.


q.e.d.


