5
$\begingroup$

Let $D$ and $E$ be toposes and let $f_{\ast}\colon D\to E$ be the direct image part of a geometric morphism $(f^{\ast},f_{\ast})$ between them. Considered as categories, we have (covariant) power-object endofunctors on each: $$P_D\colon D\to D \hspace{.5in} P_E\colon E\to E$$ where, for a morphism $\phi$ in $D$ we have $P_D(\phi)=\exists_\phi$, sending a sub-object of the domain to its image under $\phi$.

I'm trying to construct a natural transformation $$A_f\colon\ f_{\ast}\circ P_D\to P_E\circ f_{\ast}$$ of functors $D\to E\ $.

Question: For what geometric morphisms $(f^\ast,f_\ast)$ is such a natural transformation $A_f$ guaranteed to exist?

For example, such a thing exists in the case of change-of-base morphisms between slice toposes of ${\bf Set}$. If $q\colon X\to Y$ is a function, it induces a logical morphism $\Pi_q\colon {\bf Set}/X\to {\bf Set}/Y$. In this case the natural transformation $$A^~_{\Pi^~_q}\colon\Pi_q\circ P_{{\bf Set}/X}\to P_{{\bf Set}/Y}\circ \Pi_q$$ exists. It acts fiberwise on $Y$; for each $y\in Y$ it sends a $q^{-1}(y)\ $-indexed collection of subsets to their product.

How to construct this map $A_f$ in general? I wanted to use what would generalize to the morphism $f_{\ast}f^{\ast}\Omega_E\to\Omega_E\ $ induced by the mono-part of the epi-mono factorization for $\Omega_E\to f_{\ast}f^{\ast}\Omega_E$, where $\Omega_E$ is the subobject classifier in $E$. But while such a map does exist for all geometric morphisms, and can be used to construct the components of my desired $A_f$, I couldn't see how to prove that the naturality squares for $A_f$ commute. I showed it in the ${\bf Set}$ case using a basic set-theoretic argument.

So what made it work for slice toposes of ${\bf Set}$? Was it very specific to that case? Was it that these change-of-base functors are logical morphisms, or that they're essential geometric, or does such an $A_f$ always exist?

$\endgroup$

2 Answers 2

5
$\begingroup$

Since every power object is an internal Heyting algebra, and $f_*$ preserves the structure of internal Heyting algebras, there are trivial examples of such natural transformations corresponding to the constants $\top$ and $\bot$. Of course, this is uninteresting.

Let me write $P^{\mathcal{D}}$ and $P^{\mathcal{E}}$ for the respective contravariant power object functors. Since $f_*$ preserves monomorphisms, there is a canonical comparison morphism $f_* \Omega_{\mathcal{D}} \to \Omega_{\mathcal{E}}$; since $f_*$ preserves products, there is a canonical natural morphism $f_* (Y^X) \to (f_* Y)^{f_* X}$; and so there is a canonical natural morphism $f_* P^{\mathcal{D}} X \to (f_* \Omega_{\mathcal{D}})^{f_* X} \to P^{\mathcal{E}} f_* X$. So there is an interesting canonical natural transformation $\theta : f_* P^{\mathcal{D}} \Rightarrow P^{\mathcal{E}} f_*$.

Now allow me to argue using generalised elements. Let $T$ be an arbitrary object of $\mathcal{E}$, and let $p : X \to Y$ be a morphism in $\mathcal{D}$. Given a generalised element $t : T \to f_* P_\mathcal{D} X$, what is $\theta_X \circ t : T \to P_{\mathcal{E}} f_{\ast} X$, and what is $f_* \exists_p \circ t : T \to f_* P_{\mathcal{D}} Y$? Let $t' : f^* T \to P_\mathcal{D} X$ be the left adjoint transpose of $t$, and let $A' \rightarrowtail X \times f^* T$ be the subobject classified by $t'$. It is clear that $\theta_X \circ t$ is just the classifying morphism for the pullback of $f_* A' \rightarrowtail f_* X \times f_* f^* T$ along $f_* X \times T \to f_* X \times f_* f^* T$. Also, by naturality, $f_* \exists_p \circ t$ must be the right adjoint transpose of $\exists_p \circ t' : f^* T \to P_\mathcal{D} Y$, which is none other than the classifying morphism for the image of the composite $A' \rightarrowtail X \times f^* T \to Y \times f^*T$.

This suggests the crucial criterion is that $f_*$ preserve epimorphisms (and hence, epi–mono factorisations) – and this automatic for all base change morphisms for slices over $\textbf{Set}$ because $\textbf{Set}$ and its slices have the axiom of choice. So assume $f_*$ preserves epimorphisms. If we write $A \rightarrowtail f_* X \times T$ for the subobject classified by $\theta_X \circ t$, $B' \rightarrowtail Y \times f^* T$ for the image of $A' \rightarrowtail X \times f^* T \to Y \times f^* T$, and $B \rightarrowtail f_* Y \times T$ for the subobject classified by $\theta_Y \circ f_* \exists_p \circ t$, then the preservation of epi–mono factorisations implies that $f_* B'$ remains the image of $f_* A'$ under $f_* X \times f_* f^* T \to f_* Y \times f_* f^* T\ $; but epi–mono factorisations are stable under pullback in a topos, hence $B$ is the image of $A$ under $f_* X \times T \to f_* Y \times T$. Thus, we have $$\theta_Y \circ f_* \exists_p \circ t = \exists_{f_{\ast} p} \circ \theta_X \circ t$$ for all generalised elements $t : T \to f_* P_{\mathcal{D}} X$, and thus $\theta$ is also a natural transformation $f_* P_{\mathcal{D}} \Rightarrow f_* P_{\mathcal{E}}$.

$\endgroup$
2
  • $\begingroup$ That $f_\ast$ should preserve epis was the conclusion of my analysis as well (but when I came here to write it out, your answer was here already (-:). $\endgroup$
    – Todd Trimble
    Mar 17, 2013 at 1:38
  • 1
    $\begingroup$ For posterity, let $D$ be the topos of cospans in ${\bf Set}$ and let $E$ be the topos ${\bf Set}$ of sets. The unique geometric morphism $f_{\ast}\colon D\to E\ $ sends each cospan to its fiber product. It does not preserve epis, indeed let $X$ be the cospan $\{1\}\to\{1,2\}\leftarrow\{2\}\ $ and let $Y$ be the terminal cospan. The unique morphism $p\colon X\to Y\ $ is epi but $f_{\ast}(X)=\emptyset$ whereas $f_{\ast}(Y)=1\ $. One can check that the components given by $\theta$ constructed above do not form a naturality square for $p$. $\endgroup$ Mar 18, 2013 at 4:07
9
$\begingroup$

A different way to describe the same answer that Zhen and Todd arrived at is to work in the internal logic of $E$. That way we may pretend that $f_*$ is the global sections functor $\mathrm{Hom}(1,-) : D \to \mathrm{Set}$, as long as we treat $\mathrm{Set}$ constructively. Then we have the components of a putative natural transformation

$$ \mathrm{Hom}(1,P A) \to P(\mathrm{Hom}(1,A)) $$

which, under the universal property of power objects $ \mathrm{Hom}(1,P A) \cong \mathrm{Sub}(A)$, sends a subobject $S\rightarrowtail A$ in $D$ to the set of all global sections $1 \to A$ which factor through it. The naturality square for $p:A\to B$ requires that if we take the direct image subobject $p_!(S)$, then a global section of $B$ factors through $p_!(S)$ just when it lifts to some global section of $A$ factoring through $S$. It's easy to see that this is the same as asking that $1\in D$ be projective, which is equivalent to saying that the global sections functor $f_* = \mathrm{Hom}(1,-)$ preserves epimorphisms.

$\endgroup$
3
  • $\begingroup$ Looks neat, but how do we formalize our ability to "pretend that $f_{\ast}$ is the global sections functor..."? Maybe a reference for understanding a morphism $f_{\ast}\colon D\to E$ of topoi in terms of the internal logic of $E$ would help me understand this. $\endgroup$ Mar 17, 2013 at 14:39
  • 2
    $\begingroup$ If we have a geometric morphism $f : \mathcal{D} \to \mathcal{E}$, then $f^*$ makes $\mathcal{D}$ into an $\mathcal{E}$-indexed topos $\mathbb{D}$, whose fibre over an object $E$ is the slice $\mathcal{D}_{/ f^* E}$. Johnstone explains this in detail in Chapter B3 of Sketches of an elephant. A less high-tech version of this is to simply consider $\mathcal{D}$ as an $\mathcal{E}$-enriched category, with $\underline{\mathcal{D}}(X, Y) = f_*(Y^X)$; then $f_* : \mathcal{D} \to \mathcal{E}$ becomes a enriched-representable functor in an obvious way. $\endgroup$
    – Zhen Lin
    Mar 17, 2013 at 16:00
  • 1
    $\begingroup$ Now that I finally understand Zhen's answer, I realize how cool the perspective of Mike's answer is. I don't think I could have easily started there, but it looks like a very powerful approach, and I'm about to sink my teeth into Johnstone B.3 to get a piece of it. Thanks guys! $\endgroup$ Mar 18, 2013 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.