## 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.