## 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.
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