Some properties of a commutative diagram with exact rows

Consider the following commutative diagram of groups.

A commutative diagram of groups

Suppose the rows are exact; that is, \mathsf{im}\ \psi_k = \mathsf{ker}\ \varphi_k for k \in \{1,2\}. Prove the following.

  1. If \varphi_1 and \alpha are surjective and \beta is injective, then \gamma is injective.
  2. If \psi_2, \alpha, and \gamma are injective, then \beta is injective.
  3. If \varphi_1, \alpha, and \gamma are surjective, then \beta is surjective.
  4. If \beta is surjective and \gamma and \psi_2 are injective, then \alpha is surjective.

  1. Let c \in \mathsf{ker}\ \gamma. Since \varphi_1 is surjective, there exists b \in B_1 such that \varphi(b) = c. Now (\gamma \circ \varphi_1)(b) = 1, so that (\varphi_2 \circ \beta)(b) = 1, and so \varphi_2(\beta(b)) = 1. Thus \beta(b) \in \mathsf{ker}\ \varphi_2, and so \beta(b) \in \mathsf{im}\ \psi_2. Say \beta(b) = \psi_2(a^\prime) where a^\prime \in A_2. Since \alpha is surjective, there exists a \in A_1 such that \alpha(a) = a^\prime. Now \beta(b) = (\psi_2 \circ \alpha)(a), so that \beta(b) = (\beta \circ \psi_1)(a), and so \beta(b) = \beta(\psi_1(a)). Since \beta is injective, b = \psi_1(a). In particular, b \in \mathsf{im}\ \psi_1, so that b \in \mathsf{ker}\ \varphi_1. Then c = \varphi_1(b) = 1. Thus \mathsf{ker}\ \gamma = 1, and so \gamma is injective.
  2. Let b \in \mathsf{ker}\ \beta. Now \beta(b) = 1, so that (\varphi_2 \circ \beta)(b) = 1, so that (\gamma \circ \varphi_1)(b) = 1. Thus \gamma(\varphi_1(b)) = 1. Since \gamma is injective, \varphi_1(b) = 1, so that b \in \mathsf{ker}\ \varphi_1. So b \in \mathsf{im}\ \psi_1. Thus there exists a \in A_1 such that b = \psi_1(a). Now (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a)) = \beta(b) = 1, so that a \in \mathsf{ker}\ \psi_2 \circ \alpha. Since \alpha and \psi_2 are injective, \psi_2 \circ \alpha is injective, so that a = 1. Thus b = \psi_1(1) = 1, and thus \mathsf{ker}\ \beta = 1. Thus \beta is injective.
  3. Let b \in B_2. Now \varphi_2(b) \in C_2. Since \gamma and \varphi_1 are surjective, \gamma \circ \varphi_1 is surjective, so that there exists b^\prime \in B_1 such that \varphi_2(b) = (\gamma \circ \varphi_1)(b^\prime). Note that \varphi_2(b^{-1}\beta(b^\prime)) = 1, so that b^{-1} \beta(b^\prime) \in \mathsf{ker}\ \varphi_2. Thus b^{-1}\beta(b^\prime) \in \mathsf{im}\ \psi_2, and so there exists a^\prime \in A_2 such that b^{-1}\beta(b^\prime) = \psi_2(a^\prime). Since \alpha is surjective, there exists a \in A_1 such that \alpha(a) = a^\prime. Now b^{-1}\beta(b^\prime) = (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a)). Thus b = \beta(b^\prime \psi_1(a^{-1})), and so \beta is surjective.
  4. Let a \in A_2. Now \psi_2(a) \in B_2. Since \beta is surjective, there exists b^\prime \in B_1 such that \psi_2(a) = \beta(b^\prime). Now 1 = (\varphi_2 \circ \psi_2)(a) = \varphi_2(\psi_2(a)) = \varphi_2(\beta(b^\prime)) = (\varphi_2 \circ \beta)(b^\prime) = (\gamma \circ \varphi_1)(b^\prime) = \gamma(\varphi_1(b^\prime)). Since \gamma is injective, \varphi_1(b^\prime) = 1. So b^\prime \in \mathsf{ker}\ \varphi_1, and so b^\prime \in \mathsf{im}\ \psi_1. Say a^\prime \in A_1 such that b^\prime = \psi_1(a^\prime). Now \psi_2(a) = \beta(b^\prime) = \beta(\psi_1(a^\prime)) = (\beta \circ \psi_1)(a^\prime) = (\psi_2 \circ \alpha)(a^\prime) = \psi_2(\alpha(a^\prime)). Since \psi_2 is injective, a = \alpha(a^\prime). Thus \alpha is surjective.
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: