I am working through some notes, and am on a section about locally cartesian closed categories. One of the results is if $\mathcal{C}$ is local cartesian closed and has coequalisers of reflexive pairs, then all morphisms have a factorisation into a regular epic followed by a monic.
The proof seems to rely on the following:
Claim: Let the pullback functor $f^*:\mathcal{C}/B\to \mathcal{C}/A$ (induced by $f:A\to B$ in $\mathcal{C}$) have a right adjoint. Then pulling back an epic along $f$ preserves epicness.
I can't understand why this is true since an epic in $\mathcal{C}$ is just an object of $\mathcal{C}/B$, not an arrow. So even if $f^*$ preserves colimits (hence epics), it is preserving epics in $\mathcal{C}/B$, not $\mathcal{C}$. If someone could explain to me how this works that would be very helpful.
Best Answer
As mentioned in the comments to Mark's answer, this is a proof of the claim in the question with the assumption that $C$ has products and pullbacks (and no assumptions of (local) cartesian closedness). I don't know if that helps, because as they stand the hypotheses of the question do not (seem to - at least to me - EDIT : apparently Balaji Krishna assumes products in their definition of local cartesian closedness, so in this case it is sufficient) imply the existence of products.
So let $f:A\to B$ be such that $f^* : C/B\to C/A$ has a right adjoint, in particular it preserves pushouts and thus epis (epis in $C/B$, of course)
Now let $g:D\to B$ be an epi, we wish to show that $f^*g : D\to A$ is epi in $C$. To avoid confusion, I will write objects of the comma categories as couples.
Note that $g$ can also be seen as a morphism $(D,g)\to (B,id_B)$, and obviously this morphism is epi in $C/B$. Therefore, applying $f^*$ to it is still epi. Now you can check that $f^*$ of the morphism $g$ is actually still $f^*g$ (for the simple reason that $f^*(B,id_B) = (A,id_A)$ and that $f^*$(morphism $g$) is a morphism from $f^*(D,g)$ to $f^*(B,id_B)=(A,id_A)$.
Therefore $f^*g$ is epi in $C/A$, when it's seen as a morphism to $(A,id_A)$. Now we are reduced to the following claim :
(I am too lazy to draw the diagrams on here, but it's much better if you do that)
Proof : Let $l_1,l_2 : E\to T$ be arrows in $C$ such that $l_1\circ k = l_2\circ k$; and consider the object $(T\times E, h\circ \pi_E)$ of $C/A$. Then we have, for $i\in\{1,2\}$, a map $\tilde{l_i} : (E,h)\to (T\times E, h\circ \pi_E)$ : on $T$ it's $l_i$ and on $E$ it's $id_E$. It is very easy to see that it's a map of $C/A$, moreover one easily checks that $\tilde{l_1}\circ k = \tilde{l_2}\circ k$, so that by epi-ness of $k$ in $C/A$, $\tilde{l_1}=\tilde{l_2}$.
Now look at the $T$ coordinate, you get $l_1=l_2$, so $k$ is epi in $C$.
It follows that $f^*g$ is epi in $C$ and we are done.