Exercise A0.III.6.10

Proposition

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

Proof

Let $M, N$ and $Z$ be $R$-modules and let $f : M \to Z$ and $g : N \to Z$ be morphisms. We will construct the module $M \times_Z N$ with two projections $\pi_M : M \times_Z N \to M$ and $\pi_N : M \times_Z N \to N.$ We will then show that it satisfies the universal property of a fibered product of $M$ and $N$ over $Z.$

Consider the set of pairs

$$M \times_Z N = \{ (m, n) \mid m \in M, n \in N, f(m) = g(n)\}$$

with the standard projections $\pi_M : (m,n) \mapsto m,$ and $\pi_N : (m,n) \mapsto n.$

By construction, this satisfies $f \circ \pi_M = g \circ \pi_N.$

We define addition on it as

$$(m_1, n_1) + (m_2, n_2) = (m_1 + m_2, n_1 + n_2).$$

This works, since if $f(m_1) = g(n_1)$ and $f(m_2) = g(n_2)$ then

$$\begin{align*} f(m_1 + m_2) & = f(m_1) + f(m_2) \\ & = g(n_1) + g(n_2) \\ & = g(n_1 + n_2). \end{align*}$$

We also have $R$ acting on $M \times_Z N$ by

$$r(m,n) = (rm, rn).$$

This is also valid, since if $f(m) = g(n)$ then

$$f(rm)=rf(m)=rg(n)=g(rn).$$

So $M \times_Z N$ is a subset of $M \oplus N$ closed under addition and the action of $R,$ hence it is a submodule of $M \oplus N.$

Now, suppose some $R$-module $X$ has morphisms $u : X \to M$ and $v : X \to N$ such that for all $x \in X$, $f(u(x)) = g(v(x)).$ We can define a morphism $\sigma : X \to M \times_Z N$ by $x \mapsto (u(x), v(x)).$

Note that $\sigma$ is simply the restriction of the pairing $(u, v) : X \to M \oplus N$ to the submodule $M \times_Z N.$ (This makes it clear that $\sigma$ is an $R$-module homomorphism.) Indeed, $(u(x), v(x))$ has to land in $M \times_Z N$ since $f(u(x)) = g(v(x)).$

The pairing morphism is uniquely determined by $u$ and $v$, and so if there were a morphism $\sigma'$ with the same properties it would have to be the restriction of the unique pairing to the same submodule, resulting in the same morphism. That is, given the inclusion $\iota: M \times_Z N \hookrightarrow M \oplus N$ we see that

$$\iota \circ \sigma' = (u, v) = \iota \circ \sigma,$$

and since $\iota$ is a monomorphism, $\sigma' = \sigma.$

So, $\sigma$ is uniquely determined by $u$ and $v,$ which demonstrates that our construction satisfies the universal property of a fibered product.

$\blacksquare$