Morphism of topoi - the canonical topology

Question

Let $T,T'$ be Grothendieck topoi and $(f^\ast,f_\ast): T' \to T$ a morphism of topoi, i.e. $f^\ast\colon T\to T'$ is left adjoint to $f_\ast\colon T\to T'$ and $f^\ast$ commutes with finite limits.

In exercise 2.N(iii) in M. Olsson's book "Algebraic Spaces and Stacks" one is asked to show that $f^\ast$ maps a covering in the canonical topology of $T$ to a covering of the canonical topology of $T'$.

My question is: How is this proven? I tried for some time now, but I did not get far. Below I give further context and a reduction that I managed to show. Maybe I made some mistake there and that is the source of my confusion.

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Olsson works with pretopologies, so a covering in the canonical topology on an object $F\in\operatorname{obj}(T)$, may be given as a family of morphisms $\{F_i \to F\}_{i\in I}$, where $I$ is some index set such that the following diagram is an equalizer diagram for all $G\to F$ and $H$ in $T$: $$ \operatorname{Hom}_T(G,H) \to \prod_{i\in I}\operatorname{Hom}_{T}(G\times_F F_i , H) \rightrightarrows \prod_{i,j\in I}\operatorname{Hom}_{T}(G\times_F F_i \times_F F_j , H). $$ So now we want $\{f^\ast(F_i) \to f^\ast(F)\}_{i\in I}$ fulfilling the analogous criterion in $T'$, i.e. one has to show $$ \operatorname{Hom}_{T'}(G',H') \to \prod_{i\in I}\operatorname{Hom}_{T'}(G'\times_{f^\ast(F)} f^\ast(F_i) , H') \rightrightarrows \prod_{i,j\in I}\operatorname{Hom}_{T'}(G'\times_{f^\ast(F)} f^\ast(F_i) \times_{f^\ast(F)} f^\ast(F_j) , H'), $$ for every $G\to f^\ast(F)$ and $H$ in $T'$, is exact. If one assumes that $G'=f^\ast(G)$ for some $G\in\operatorname{obj}(T)$ and that $G'\to f^\ast(F)$ comes from some $G\to F$ adjointsness and commutativity of limits yields the desired result, but in the general case ...

If I assume that $T'=\operatorname{Shv}(\mathcal{C})$ for some site $\mathcal{C}$, I can reduce to the case that $G'$ is representable (i.e. the sheafification of $\operatorname{Hom}_{T'}(-,V)$ for some $V\in\operatorname{obj}(T')$), since given any family of morphisms $$ \{ \varphi_i \colon G' \times_{f^\ast(F)} f^\ast(F_i) \to H' \}_{i\in I} $$ that fulfill the equalizing condition, we can construct a $\varphi\colon G'\to H'$ via:

Taking $g\in G'(V)$ for $V\in\operatorname{obj}(\mathcal{C})$ we apply the assumption on representables to the family $$ \{ \operatorname{Hom}_{T'}(-,V)^a \times_{f^\ast(F)} f^\ast(F_i) \to G' \times_{f^\ast(F)} f^\ast(F_i) \to H'\}, $$ where the superscript $a$ is supposed to denote sheafification and $\operatorname{Hom}_{T'}(-,V) \to G'$ is induced by $g$ via Yoneda's lemma. Thus we get a morphism $\operatorname{Hom}_{T'}(-,V)^a \to H'$ or equivalently an element $\varphi(g)\in H'(V)$. We use this to construct $\varphi\colon G' \to H'$. But is this even a step in the right direction?

It feels to me that this is some standard problem, but I could not find a solution. Any nudge in the right direction helps.

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Thanks in advance and have a great day!

Answer 1

It seems that your question is resolved as soon as you are able to identify the canonical topology on a Grothendieck topos, so let me state it explicitly.

Proposition. The canonical topology on a Grothendieck topos has as its covering families all small jointly epimorphic sinks.

As you surmised, this is because epimorphisms in a topos are effective and stable under pullback; in other words, in a topos, epimorphism = universal effective epimorphism.

Your original question about the inverse image functor is now easy to answer: left adjoints preserve colimits, so small jointly epimorphic sinks are sent to small jointly epimorphic sinks; in other words, they preserve coverings w.r.t. the canonical topology.

Answer 2

This is basically what @ZhenLin explained above, I just want to summarize it to close this question.

(EDIT: @ZhenLin posted his answer right before mine, so I just leave this up to not let the effort go to waste.)

The definition for a covering in the canonical topology that was given in the question is general and applies to any category with fiber products. In the case that the underlying category is a topos $T$, i.e. the category of sheaves for some site $\mathcal{C}$ one finds, that for every $F\in\operatorname{obj}(T)$ a covering of $F$ is already given by a family $\{F_i \to F\}_{i\in I}$ which is jointly epimorphic. Since small coproducts are available in $T$ this is equivalent to $\bigsqcup_{i\in I} F_i \to F$ being an epimorphism.

Now since epimorphisms are preserved by any left adjoint, we arrive at the claim of the exercise.

I just want to fix a proof for the transition from the definition of a covering in the canonical topology as in the question to the characterization above, because that is what kept me from seeing this in the first place.

So given an object $F\in\operatorname{obj}(T)$ and a jointly epimorphic family $\{F_i \to F\}_{i\in I}$ in $T$, we show that $$ \operatorname{Hom}_T(G,H) \to \prod_{i\in I} \operatorname{Hom}(G\times_F F_i,H) \rightrightarrows \prod_{i,j\in I} \operatorname{Hom}(G\times_F F_i \times_F F_j , H) $$ is exact, for every $G\to F$ and $H$ in $T$.

Lemma 1: Every epimorphism $\varphi\colon K\to L$ in $T$ is an effective epimorphism, i.e. $$ K \times_L K \rightrightarrows K \xrightarrow{\varphi} L $$ is exact, where the two arrows on the left are the canonical projections.

Proof: In order to show the claim, we have to check that for every $Z\in\operatorname{obj}(T)$ the following sequence is exact: $$ \operatorname{Hom}_T(L,Z) \hookrightarrow \operatorname{Hom}_T(K,Z) \rightrightarrows \operatorname{Hom}(K\times_L K,Z). $$ As $K\to L$ is an epimorphisms the $\operatorname{Hom}_T(L,Z) \to \operatorname{Hom}_T(K,Z)$ is indeed a monomorphism. And any $L\to Z$ fits into a diagram $$ \require{AMScd} \begin{CD} K \times_L K @>\operatorname{pr}_2>> K \\ @V\operatorname{pr}_1VV @VV{\varphi}V \\ K @>>{\varphi}> L @>>> Z. \end{CD} $$ So let $\psi\colon K\to Z$ be a morphism such that $\psi\circ \operatorname{pr}_1 = \psi\circ\operatorname{pr}_2$ holds. Now we construct a $\psi'\colon L\to Z$ such that $\psi = \psi' \circ \varphi$. We therefore use that $T$ is the category of sheaves over the site $\mathcal{C}$.

Take $C \in \operatorname{obj}(\mathcal{C})$ and $f\in L(C)$. Since $\varphi$ is an epimorphism there is a covering $\{C_i \to C\}_{i\in I}$ in $\mathcal{C}$ and $g_i\in K(C_i)$ such that $\varphi(g_i) = f_{\restriction C_i}$. From $\psi\circ \operatorname{pr}_1 = \psi\circ\operatorname{pr}_2$ we can deduce that whenever $\varphi(g_1) = \varphi(g_2)$ for some $g_1,g_2\in K(C')$ with $C'\in\operatorname{obj}(\mathcal{C})$ holds, $\psi(g_1)=\psi(g_2)$ follows. Since we have $$ \varphi(g_{i\restriction C_i \times_C C_j}) = f_{\restriction C_i\times_C C_j} = \varphi(g_{j\restriction C_i \times_C C_j}) $$ we get $$ \psi(g_{i\restriction C_i \times_C C_j}) = \psi(g_{j\restriction C_i \times_C C_j}) $$ for all $i,j \in I$. Since $Z$ is a sheaf, we can glue the $\psi(g_i)$ to obtain an element $\psi'(f)\in Z(C)$. Modulo checking naturality and well-definedness this gives the desired morphism $\psi'\colon L\to Z$.

Lemma 2: Epimorphisms in $T$ are stable under pullbacks.

Proof: Let $\varphi\colon K\to L$ be an epimorphism and $\xi\colon M\to L$ some morphism in $T$. We want to check that $\varphi'\colon K\times_L M \to M$ is an epimorphism, so fix $\lambda_1,\lambda_2\colon M\to M'$ with $\lambda_1 \circ \varphi' = \lambda_2 \circ \varphi'$.

Let there a $C\in\operatorname{obj}(\mathcal{C})$ and some $f\in M(C)$. We set $f':=\xi(f)$. Since $\varphi$ is an epimorphism of sheaves there is a covering $\{C_i\to C\}_{i\in I}$ in $\mathcal{C}$ and $g_i\in K(C_i)$ such that $\varphi(g_i) =f'_{\restriction C_i}$ for all $i\in I$. Now we may view $(g_i,f_{\restriction C_i})$ as an element of $(K\times_L M)(C_i)$, whence $\lambda_1(f_{\restriction C_i}) = \lambda_2(f_{\restriction C_i})$ for all $i\in I$. Since $M'$ is a sheaf $\lambda_1(f)=\lambda_2(f)$ follows and thus $\varphi'$ is an epimorphism.

Now we can conclude: As $\{F_i\to F\}_{i\in I}$ is jointly epimorphic, we have that $\bigsqcup_{i\in I} F_i \to F$ is an epimorphism, by Lemma 1 and Lemma 2 this morphism is even a universally effective epimorphism. But this is just the definition of a covering of $F$ in the canonical topology given in the question.