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 $b \circ f = f'$.

  • 2
    $\begingroup$ Yay you avoided an X Y type problem by saying what you were trying to do at first. So satisfactory. $\endgroup$ Commented May 17 at 19:37

1 Answer 1


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 @<g<<P \\ @VbVV @VcVV @| \\ B' @>\pi'>> C' @<g'<< P \\ \end{CD} 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.

  • $\begingroup$ 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. $\endgroup$
    – George
    Commented May 17 at 19:08
  • 1
    $\begingroup$ @George I've edited my answer to address your underlying question about projectivity in the arrow category. $\endgroup$ Commented May 17 at 19:19
  • $\begingroup$ Ooh, defining it that way makes a lot of sense! Can't believe I missed that. Thank you, I was just starting to write up a question for that situation - I would upvote again if I could! $\endgroup$
    – George
    Commented May 17 at 19:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .