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.
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: