Exercise A0.III.6.11

Proposition

Pushouts exist in $R\mathrm{-Mod}.$

Proof

Let $M, N$ and $A$ be $R$-modules and let $f : A \to M$ and $g : A \to N$ be morphisms. We will construct the module $M \oplus_A N$ with two injections $\iota_M : M \to M \oplus_A N$ and $\iota_N : N \to M \oplus_A N.$ We will then show that it satisfies the universal property of a fibered coproduct of $M$ and $N$ over $A.$

First, consider the direct sum $M \oplus N$ (we will implicitly identify it with the cartesian product for convenience). Let $U \subseteq M \oplus N$ be a subset of $M \oplus N$ given as

$$U := \left\{ \left(f\left(a\right), - g\left(a\right)\right) \mid a \in A\right\}.$$

We can verify that $U$ is a submodule by noting that, given arbitrary $r \in R$ and $a, b \in A,$ we have

$$r(f(a), -g(a)) = (rf(a),-rg(a)) = (f(ra),-g(ra)),$$

and

$$\begin{align*} (f(a),-g(a)) + (f(b),-g(b)) &= (f(a)+f(b),-(g(a)+g(b))) \\ & = (f(a+b),-g(a+b)). \end{align*}$$

Let $\pi : M \oplus N \to (M \oplus N)/U$ and define $\iota_M : M \to (M \oplus N)/U$ as $m \mapsto \pi(m,0).$ Similarly, define $\iota_N : N \to (M \oplus N)/U$ as $n \mapsto \pi(0,n).$

We can easily check that $\iota_M \circ f = \iota_N \circ g.$ Let $a \in A$ and consider

$$\begin{align*} \iota_M(f(a)) &= \pi(f(a),0) \\ &= (f(a),0) + U \\ &= (f(a),0) - (f(a),-g(a)) + U \\ &= (0,g(a)) + U \\ &= \pi(0,g(a)) \\ &= \iota_N(g(a)). \end{align*}$$

We will now show that $(M \oplus N)/U$ satisfies the universal property of a pushout. Suppose that $h_1 : M \to X$ and $h_2 : N \to X$ are $R$-module homomorphisms and furthermore $h_1 \circ f = h_2 \circ g.$ We want to find the morphism $\sigma : (M \oplus N)/U \to X,$ uniquely fixed by our data.

Consider the copairing $[h_1, h_2] : M \oplus N \to X.$ It is fixed by $h_1$ and $h_2.$ We can be certain that $U \subseteq \ker\left(\left[h_1,h_2\right]\right),$ since for any $a \in A,$

$$\begin{align*} [h_1,h_2](f(a),-g(a)) &= h_1(f(a))-h_2(g(a))\\ &= 0. \end{align*}$$

And thus, by the universal property of the quotient, $[h_1,h_2]$ factors uniquely as $\sigma \circ \pi$ for some morphism $\sigma : (M \oplus N)/U \to X.$

Does $\sigma$ make the diagram commute? We can take $m \in M$ and check as follows

$$\begin{align*} \sigma(\iota_M(m)) &= \sigma(\pi(m,0)) \\ & =[h_1,h_2](m,0) \\ & =h_1(m), \\ \end{align*}$$

and a symmetric argument works for the $N$ side of the diagram.

Is $\sigma$ unique? Take $\tau : (M \oplus N)/U \to X$ and note that, since the direct sum injections $i_M, i_N$ are jointly epimorphic, the constraints

$$\sigma \circ \pi \circ i_M = \tau \circ \pi \circ i_M \quad \text{and} \quad \sigma \circ \pi \circ i_N = \tau \circ \pi \circ i_N$$

force $\sigma \circ \pi = \tau \circ \pi.$ Since $\pi$ is itself an epimorphism (by canonical decomposition), $\sigma \circ \pi = \tau \circ \pi$ implies $\sigma = \tau.$

This demonstrates that $(M \oplus N)/U$ is indeed the fibered coproduct $M \oplus_A N,$ as required.

$\blacksquare$