5
$\begingroup$

Let $C$ be a regular category with pushouts and $S(X)$ is the lattice of subjects of $X$. For every arrow $f\colon X\to A$, pulling back along $f$ gives a map $f^*\colon S(A)\to S(X)$ which has a left adjoint $f_*$. These maps satisfy the Beck-Chevalley condition on a square $fg=hk$ if $h^*f_*=k_*g^*$.

The BC condition holds on all pullbacks. Does it hold also on pushouts? If no, do you know a class of regular categories where this is true (exact, abelian, pretoposes...)?

$\endgroup$
4
$\begingroup$

No, the Beck-Chevalley condition does not hold for all pushout squares in a regular category, and not even if the category is exact, or a pretopos, or even a topos. In fact, here is a counterexample in $\rm Set$. Let $B = \{a,b\}$ and $C = \{\alpha,\beta\}$ and $A=\{0,1,2\}$, define $f:A\to B$ by $f(0)=f(1)=a$ and $f(2)=b$, and $g:A\to C$ by $g(0)=\alpha$ and $g(1)=g(2)=\beta$. Let $P$ be the pushout of $f$ and $g$ with coprojections $p:B\to P$ and $q:C\to P$. Then

$$q(\alpha) = q(g(0)) = p(f(0)) = p(a) = p(f(1)) = q(g(1)) = q(\beta) = q(g(2)) = p(f(2)) = p(b)$$

so $P$ is a one-element set. This, if we let $U=\{b\} \in S(B)$, then $p_!(U)$ is all of $P$, hence $q^* p_!(U)$ is all of $C$. (The notation $p_!$ for left adjoints of $p^*$ is, I think, more common than $p_*$ in this field, the latter being used more often for right adjoints.) But $f^*(U) = \{2\}$, so $g_! f^*(U) = \{\beta\} \neq q^* p_!(U)$.

There is something positive that can be said, however. Suppose $P = B \amalg_A C$ is a pushout of $f:A\to B$ and $g:A\to C$ such that the induced map $\Delta : A \to B\times_P C$ is a regular epimorphism. Then the pullback square defining $B\times_P C$, consisting of $h:B\times_P C\to B$ and $k:B\times_P C\to C$ with $p:B\to P$ and $q:C\to P$, does satisfy the Beck-Chevalley condition. Moreover, since $\Delta$ is a regular epimorphism, we have $\Delta_! \Delta^* = \rm Id$, and therefore

$$ g_! f^* = k_! \Delta_! \Delta^* h^* = k_! h^* = q^* p_! $$

so the original pushout square does satisfy the Beck-Chevalley condition.

The above "zig-zag" counterexample is simply the "minimal" situation in which $\Delta$ fails to be surjective. (One might call it "the minimal way to violate the hypotheses of the baby Blakers-Massey theorem".) Thus, it seems that $\Delta$ being a regular epi is probably the best possible hypothesis.

$\endgroup$
  • $\begingroup$ Dear Mike, thank you for tour answer, it is very useful. $\endgroup$ – Gir Apr 11 '19 at 8:10
  • $\begingroup$ Is there a better name for coprojections? I know it makes sense, but the canonical inclusions into a sum aren't called coprojections. $\endgroup$ – Andrej Bauer Apr 11 '19 at 10:06
  • 2
    $\begingroup$ @AndrejBauer They aren't? I call them that. One problem with "inclusion" or "injection" is, of course, that they aren't in general injective. I'm generally in favor of using explicitly dual terminology for things that are in fact dual. $\endgroup$ – Mike Shulman Apr 11 '19 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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