$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\PHom{PHom}$I'm reading about stable module categories, and I have a question about the definition of the maps. Let $R$ be a ring, and take (left) modules $M$ and $N$. Then let $f,g\in \Hom_R(M,N)$. We then say that $f\sim g$ if $f-g\in \PHom_R(M,N)$, where: $$\PHom_R(M,N) := \{ f\in \Hom_R(M,N)\mid f \text{ factors through a projective}\}.$$

Now, this is an equivalence relation, but I'm not sure how the transitivity works. Consider $f\sim g$ and $g\sim h$. Then $f-g = \ell_1\circ k_1$ and $g-h = \ell_2\circ k_2$ (where $k_i \in \Hom_R(M,P_i)$ and $\ell_i\in \Hom_R(P_i,N)$, for $P_i$ projective). I now want to find a new projective module, which I will call $Q$ such that $f-h$ factors through it. However, I cannot tell how this occurs, as the mappings that we have do not seem to work well with the type of composition that we are after. It is also entirely unclear how one would come up with the module $Q$.

Any insight would be appreciated. I think I'm just missing one crucial thing.


1 Answer 1


This equivalence relation is just a quotient of abelian groups. It will be cleaner to show that the $\mathrm{PHom}_R(M, N)$ is a subgroup of $\mathrm{Hom}_R(M,N).$ Indeed, if $f:M\to P_1 \to N$, $g: M \to P_2\to N$ are two maps, we may add them together as $$f+g:M\to P_1\oplus P_2\to N.$$

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    $\begingroup$ That is cleaner, but it's not much less clean to do it directly for equivalence relations: One just has to replace $f$ and $g$ in your formulation with $f-g$ and $g-h$, and use $(f-g)+(g-h)=(f-h)$. $\endgroup$
    – Will Sawin
    Apr 12 at 0:09
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    $\begingroup$ Thank you very much! You are absolutely correct, that is much cleaner! $\endgroup$ Apr 14 at 16:11

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