Lemma 15.70.1. Let $R$ be a ring. Let $M$, $N$ be $R$-modules.

1. Given an $R$-module map $\varphi : M \to N$ the following are equivalent: (a) $\varphi$ factors through a projective $R$-module, and (b) $\varphi$ factors through a free $R$-module.

2. The set of $\varphi : M \to N$ satisfying the equivalent conditions of (1) is an $R$-submodule of $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$.

3. Given maps $\psi : M' \to M$ and $\xi : N \to N'$, if $\varphi : M \to N$ satisfies the equivalent conditions of (1), then $\xi \circ \varphi \circ \psi : M' \to N'$ does too.

Proof. The equivalence of (1)(a) and (1)(b) follows from Algebra, Lemma 10.77.2. If $\varphi : M \to N$ and $\varphi ' : M \to N$ factor through the modules $P$ and $P'$ then $\varphi + \varphi '$ factors through $P \oplus P'$ and $\lambda \varphi$ factors through $P$ for all $\lambda \in R$. This proves (2). If $\varphi : M \to N$ factors through the module $P$ and $\psi$ and $\xi$ are as in (3), then $\xi \circ \varphi \circ \psi$ factors through $P$. This proves (3). $\square$

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