Compatibility of pullbacks with an equivalence relation

97 Views Asked by At

I'm currently working on the proof of the existence of the sheafification in Notes on Grothendieck topologies, fibered categories and descent theory , 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: $$ a,b\in F(U):a\sim b\Rightarrow f^*(a)=f^*(b) $$ I tried proving this statement with 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, $pr_1^*\circ f^*(a)=pr_2^*\circ\phi^*(a)=pr_2^*\circ\phi^*(b)=pr_1^*\circ f^*(b).$

Now the questions:

  1. Is $pr_1^*$ monic or is it already proven?
  2. Is there a different way to proof this? If yes, i just need a starting point.