## Pullbacks and pushouts of modules

Let $R$ be a ring with 1 and let $M$, $A$, and $B$ be left unital $R$-modules.

1. Let $\varphi : A \rightarrow M$ and $\psi : B \rightarrow M$ be module homomorphisms. Prove that there exists a module $X$ and two module homomorphisms $\lambda_1 : X \rightarrow A$ and $\lambda_2 : X \rightarrow B$ such that $\varphi \circ \lambda_1 = \psi \circ \lambda_2$ and which have the following property: if $Z$ is a module and $\alpha_1 : Z \rightarrow A$ and $\alpha_2 : Z \rightarrow B$ are module homomorphisms such that $\varphi \circ \alpha_1 = \psi \circ \alpha_2$, then there exists a unique module homomorphism $\theta : Z \rightarrow X$ such that $\alpha_1 = \lambda_1 \circ \theta$ and $\alpha_2 = \lambda_2 \circ \theta$. That is, given the following diagram of modules,

The pullback of two modules

there exists a unique $\theta$ which makes the diagram commute. Deduce that $X$ is unique up to isomorphism. We will call this $X$ the pullback or fiber product of $\varphi$ and $\psi$, and sometimes denote it by $\mathsf{pb}(\varphi,\psi)$.

2. Let $\varphi : M \rightarrow A$ and $\psi : M \rightarrow B$ be module homomorphisms. Prove that there exists a module $X$ and two module homomorphisms $\sigma_1 : A \rightarrow X$ and $\sigma_2 : B \rightarrow X$ such that $\sigma_1 \circ \varphi = \sigma_2 \circ \psi$ and which have the following property: If $Z$ is a module and $\alpha_1 : A \rightarrow Z$ and $\alpha_2 : B \rightarrow Z$ are module homomorphisms such that $\alpha_1 \circ \varphi = \alpha_2 \circ \psi$, then there exists a unique module homomorphism $\theta : X \rightarrow Z$ such that $\theta \circ \sigma_1 = \alpha_1$ and $\theta \circ \sigma_2 = \alpha_2$. That is, given the following diagram of modules,

The pushout of two module homomorphisms

there exists a unique $\theta$ which makes the diagram commute. Deduce that $X$ is unique up to isomorphism. We will call this $X$ the pushout or fiber sum of $\varphi$ and $\varphi$ and $\psi$, and sometimes denote it by $\mathsf{po}(\varphi,\psi)$.

1. Let $X = \{ (a,b) \in A \times B \ |\ \varphi(a) = \psi(b) \}$, and let $\lambda_1(a,b) = a$ and $\lambda_2(a,b) = b$. Note that if $(a_1,b_1), (a_2,b_2) \in X$ and $r \in R$, then $\varphi(a_1+ra_2) = \psi(b_1+rb_2)$, so that $(a_1+ra_2,b_1+rb_2) \in X$. Since $(0,0) \in X$, by the submodule criterion, $X$ is a submodule of $A \times B$. Certainly if $(a,b) \in X$, then $\varphi(a) = \psi(b)$, so that $(\varphi \circ \lambda_1)(a,b) = (\psi \circ \lambda_2)(a,b)$. Thus $\varphi \circ \lambda_1 = \psi \circ \lambda_2$. Now suppose we have $Z$, $\alpha_1$, and $\alpha_2$. Note that for all $z \in Z$, we have $\varphi(\alpha_1(z)) = \psi(\alpha_2(z))$, so that $(\alpha_1(z), \alpha_2(z)) \in X$. Define $\theta : Z \rightarrow X$ by $\theta(z) = (\alpha_1(z), \alpha_2(z))$. Clearly $\theta$ is a module homomorphism. Moreover, $\lambda_1 \circ \theta = \alpha_1$ and $\lambda_2 \circ \theta = \alpha_2$. The uniqueness of $\theta$ follows easily.
2. Define $Y = \{ (\varphi(m), -\psi(m)) \ |\ m \in M \} \subseteq A \oplus B$. Note that $(0,0) \in Y$, and if $(\varphi(m_1),-\psi(m_1)), (\varphi(m_2), -\psi(m_2)) \in Y$ and $r \in R$, then $(\varphi(m_1),-\psi(m_1)) + r(\varphi(m_2), -\psi(m_2)) = (\varphi(m_1+rm_2), -\psi(m_1+rm_2)) \in Y$. By the submodule criterion, $Y \subseteq A \oplus B$ is a submodule. Let $X = (A \oplus B)/Y$, and define $\sigma_1 : A \rightarrow X$ and $\sigma_2 : B \rightarrow X$ by $\sigma_1(a) = (a,0) + Y$ and $\sigma_2(b) = (0,b) + Y$. Note that for all $m \in M$, we have $(\varphi(m),0) + Y = (\varphi(m),0) - (\varphi(m),-\psi(m)) + Y$ $= (0,\psi(m)) + Y$. Thus $\sigma_1 \circ \varphi = \sigma_2 \circ \psi$. Now suppose we have $Z$, $\alpha_1 : A \rightarrow Z$, and $\alpha_2 : B \rightarrow Z$. Define $\theta : X \rightarrow Z$ by $\theta((a,b)+Y) = \alpha_1(a) + \alpha_2(b)$. To see that $\theta$ is well-defined, suppose $(a_1,b_1) - (a_2,b_2) \in Y$. Then we have $a_1-a_2 = \varphi(m)$ and $b_1-b_2 = \psi(m)$ for some $m \in M$. Applying $\alpha_1$ and $\alpha_2$, we see that $\alpha_1(a_1) - \alpha_1(a_2) = (\alpha_1 \circ \varphi)(m)$ $= (\alpha_2 \circ \psi)(m) = \alpha_2(b_1) - \alpha_2(b_2)$. Thus $\alpha_1(a_1) + \alpha_2(b_1) = \alpha_1(a_2) + \alpha_2(b_2)$. Clearly $\theta$ is a module homomorphism. Moreover, we see that $\theta \circ \sigma_1 = \alpha_1$ and $\theta \circ \sigma_2 = \alpha_2$. To see uniqueness, suppose there exists $\zeta : X \rightarrow Z$ such that $\zeta \circ \sigma_1 = \alpha_1$ and $\zeta \circ \sigma_2 = \alpha_2$. Now $\zeta((a,b)+Y) = \zeta((a,0)+Y) + \zeta((0,b)+Y)$ $= (\zeta \circ \sigma_1)(a) + (\zeta \circ \sigma_2)(b)$ $= \alpha_1(a) + \alpha_2(b) = \theta((a,b)+Y)$, so that $\zeta = \theta$.