# Commutativity of second square when first square and outer rectangle commute

The general question I reach is with the diagram below $$\require{AMScd}$$ $$\begin{CD} A @>f>> B @>\pi>> C \\ @| @VbVV @VcVV \\ A' @>f'>> B' @>\pi'>> C' \\ \end{CD}$$ where the outer rectangle and the right square commute, and $$\pi$$ and $$\pi'$$ are surjections. Is this enough to guarantee that the left square commutes? I feel like not since we need left-inverses to $$\pi$$ (for which injectivity is required).

More specifically, related to "Two subcategories" part of this question on the morphism category $$\mathrm{Mor}({A})$$, I'm trying to show $$P \xrightarrow{\mathrm{id}_P}P$$ is projective for $$P \in A$$ projective, and reached $$\begin{CD} P @>f>> B @>\pi>> C \\ @| @VbVV @VcVV \\ P @>f'>> B' @>\pi'>> C' \\ \end{CD}$$ with $$\pi,\pi'$$ epic and commutativity of the right square and outer rectangle. This was obtained with $$f$$ morphisms from each triangle in $$A$$ $$\begin{CD} @. B \\ @. @V\pi VV\\ P @>g>>C \\ \end{CD}$$ since epimorphisms in $$\textrm{Mor}(A)$$ are component-wise epimorphisms (inheriting pointwise (co)limits as a functor category $$\mathrm{Fun}(* \to *, A)$$), so to finish with $$f$$ a $$\textrm{Mor}( A)$$-morphism, the left square should commute, that is $$b \circ f = f'$$.

• Yay you avoided an X Y type problem by saying what you were trying to do at first. So satisfactory. Commented May 17 at 19:37

No. If $$C'$$ is the terminal object, then the outer rectangle and right square always commute, so it's easy to cook up a counterexample in which the left square fails to commute.

For an explicit counterexample in $$\mathsf{Set}$$, let $$A=\{a\}$$, $$B=B'=\{b,b'\}$$ and $$C=C'=\{c\}$$. Let $$f(a)=b$$, $$f'(a)=b'$$, $$b=\mathrm{id}$$, and take $$\pi,\pi',c$$ all to be constant maps with value $$c$$. Then $$b\circ f=f\neq f'$$, but the rectangle and right square commute, and $$\pi,\pi'$$ are surjections.

But the thing you want to prove is true (with some mild hypotheses on the base category). Suppose $$P$$ is a projective object in a category $$\mathcal{C}$$. Let $$\mathrm{Mor}(\mathcal{C})$$ be the arrow category. Assume that whenever $$(\pi,\pi')$$ is an epimorphism in $$\mathrm{Mor}(\mathcal{C})$$, $$\pi$$ is an epimorphism in $$\mathcal{C}$$. Then $$P\xrightarrow{\mathrm{id}_P}P$$ is a projective object in $$\mathrm{Mor}(\mathcal{C})$$.

To see this, suppose $$B\xrightarrow{b}B'$$ and $$C\xrightarrow{C}C'$$ are objects, and we have an epimorphism $$(\pi,\pi')\colon b\to c$$ and a morphism $$(g,g')\colon \mathrm{id}_P\to c$$ given by the following commutative diagram:

$$\begin{CD} B @>\pi>> C @\pi'>> C' @ Since $$\pi$$ is an epimorphism in $$\mathcal{C}$$ and $$P$$ is projective, there is a morphism $$f\colon P\to B$$ such that $$\pi\circ f = g$$. Now define $$f'\colon P\to B'$$ by $$f' = b\circ f$$. Then $$(f,f')\colon \mathrm{id}_P\to b$$ is a morphism in $$\mathrm{Mor}(\mathcal{C})$$. It remains to check that $$(\pi,\pi')\circ(f,f') = (g,g')$$. Indeed, $$\pi\circ f = g$$, and $$\pi'\circ f' = \pi'\circ b\circ f = c\circ \pi\circ f = c\circ g = g'$$, as desired.

Now what about the hypothesis that whenever $$(\pi,\pi')$$ is an epimorphism in $$\mathrm{Mor}(\mathcal{C})$$, $$\pi$$ is an epimorphism in $$\mathcal{C}$$? This holds under very mild hypotheses on $$\mathcal{C}$$. For example, if $$\mathcal{C}$$ has a terminal object, it's true.

Suppose $$(\pi,\pi')\colon b\to c$$ is an epimorphism in $$\mathrm{Mor}(\mathcal{C})$$ (with $$b\colon B\to B'$$ and $$c\colon C\to C'$$). To show $$\pi$$ is an epimorphism in $$\mathcal{C}$$, suppose we have $$f,g\colon B\to D$$ such that $$f\circ \pi = g\circ \pi$$. Let $$!_D\colon D\to 1$$ be the unique arrow to the terminal object, and similarly for $$!_{C'}$$, etc. Then $$(f,!_{C'})$$ and $$(g,!_{C'})$$ are arrows in $$\mathrm{Mor}(\mathcal{C})$$ and $$(f,!_{C'}) = (f\circ \pi,!_{C'}\circ \pi') = (g\circ \pi,!_{C'}\circ \pi') = (g,!_{C'})\circ (\pi,\pi')$$. Since $$(\pi,\pi')$$ is an epimorphism, $$(f,!_{C'}) = (g,!_{C'})$$, so $$f = g$$. Thus $$\pi$$ is an epimorphism.

• Thanks, this helps a lot - this generalizes easily to the R-module case by taking those as $\mathbb Z$-module generators, which then works for $A$ projective (in the second diagram take $P=\mathbb Z,B=B'=\mathbb Z^2,C=C'=0$ and then $f$ and $f'$ inclusion into first and second coordinate respectively) which shows I need to do more than apply the maps given from each $P$ being projective. Commented May 17 at 19:08