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Let $\mathscr C$ be a cartesian category with subobject classifier $\Omega$ and generic subobject $\tau:I\to\Omega$. Consider the following pullback square:

$\require{AMScd} \begin{CD} P @>{\bar f}>> C\\ @V{\bar g}VV @V{g}VV \\ A @>{f}>> B \end{CD}$

I proved that for any $p:A\to\Omega$ the statement $\exists_{\bar f}(p\circ\bar g)\implies \exists_f(p)\circ g$ is true.

What about the converse: $\exists_f(p)\circ g\implies\exists_{\bar f}(p\circ\bar g)$?

My attempts to prove it in such a category fails. On the other hand, it seems to be intuitively true in a topos.

What assumptions are necessary in order to make it true?

Note that a similar question arises for the universal quantificator. In particular, $\forall_f(p)\circ g\implies\forall_{\bar f}(p\circ\bar g)$ is true. On the other hand, I found a proof for the converse $\forall_{\bar f}(p\circ\bar g)\implies \forall_f(p)\circ g$ when $g$ is monic.

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Is $\bar{g} p$ supposed to be $p \circ \bar{g}$ here? –  Zhen Lin May 2 at 13:13

1 Answer 1

up vote 2 down vote accepted

The equivalence $\exists_{\bar f}(p\circ\bar g)\iff \exists_f(p)\circ g$ means that the pullback preserve images. Thus that equivalence holds if and only if the categor is regular.

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