Some properties of a commutative diagram of groups with exact rows

Consider the following commutative diagram of groups.

A commutative diagram of groups

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

  1. If \alpha is surjective and \beta and \delta are injective, then \gamma is injective.
  2. If \delta is injective and \alpha and \gamma are surjective, then \beta is surjective.

  1. Let c \in \mathsf{ker}\ \gamma. Note that \chi_2(\gamma(c)) = 1, so that 1 = (\chi_2 \circ \gamma)(c) = (\delta \circ \chi_1)(c) = \delta(\chi_1(c)). Since \delta is injective, \chi_1(c) = 1. Thus c \in \mathsf{ker}\ \chi_1, and so c \in \mathsf{im}\ \varphi_1. Say b \in B_1 such that c = \varphi_1(b). Now 1 = \gamma(c)= \gamma(\varphi_1(b)) = (\gamma \circ \varphi_1)(b) = (\varphi_2 \circ \beta)(b) = \varphi_2(\beta(b)), so that \beta(b) \in \mathsf{ker}\ \varphi_2. Thus \beta(b) \in \mathsf{im}\ \psi_2. Say a^\prime \in A_2 with \beta(b) = \psi_2(a^\prime). Since \alpha is surjective, there exists a \in A_1 such that \alpha(a) = a^\prime. Now \beta(b) = (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a)). Since \beta is injective, we have b = \psi_1(a). In particular, b \in \mathsf{im}\ \psi_1, so that b \in \mathsf{ker}\ \varphi_1. Thus c = \varphi_1(b) = 1. So \mathsf{ker}\ \gamma = 1, and so \gamma is injective.
  2. Let b \in B_2. Now \varphi_2(b) \in C_2. Since \gamma is surjective, there exists c^\prime \in C_1 such that \gamma(c^\prime) = \varphi_2(b). Now \gamma(c^\prime) \in \mathsf{im}\ \varphi_2, so that \gamma(c^\prime) \in \mathsf{ker}\ \chi_2. Now (\chi_2 \circ \gamma)(c^\prime) = 1, so that (\delta \circ \chi_1)(c^\prime) = 1. Since \delta is injective, \chi_1(c^\prime) = 1. Thus c^\prime \in \mathsf{ker}\ \chi_1. So c^\prime \in \mathsf{im}\ \varphi_1. Say b^\prime \in B_1 such that \varphi_1(b^\prime) = c^\prime. Now \varphi_2(b) = \gamma(c^\prime) = (\gamma \circ \varphi_1)(b^\prime) = (\varphi_2 \circ \beta)(b^\prime) = \varphi_1(\beta(b^\prime)). Thus \varphi_2(b^{-1}\beta(b^\prime)) = 1, so that b^{-1}\beta(b^\prime) \in \mathsf{ker}\ \varphi_2. Now b^{-1}\beta(b^\prime) \in \mathsf{im}\ \psi_2; say a \in A_2 such that b^{-1}\beta(b^\prime) = \psi_2(a). Since \alpha is surjective, there exists a^\prime \in A_1 such that \alpha(a^\prime) = a. Now b^{-1}\beta(b^\prime) = (\psi_2 \circ \alpha)(a^\prime) = (\beta \circ \psi_1)(a^\prime). Thus b^{-1}\beta(b^\prime) = \beta(\psi_1(a^\prime)), and so b = \beta(b^\prime \psi_1((a^\prime)^{-1})). Thus \beta is surjective.
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