Assume I have the following commutative diagram of abelian groups, which is exact. $\require{AMScd}$ \begin{CD} A @>\psi>> B @>\varphi>>C @>\pi>> D \\ @V \alpha V V @VV \beta V @VV \gamma V @VV \delta V\\ A' @>>\psi'> B'@>>\varphi'> C' @>>\pi'> D' \end{CD}
Assume further that we know that $\alpha$ is surjective, and $\beta$ and $\delta$ are injective.
I want to show that it follows that $\gamma$ is injective.
Here is my reasoning (where we note that the diagram commutes, and is exact): I want to show that $$\gamma(c) = 0 \implies c = 0.$$
Assume $$\gamma(c) = 0 \implies (\pi' \circ \gamma)(c) = 0 \quad(\text{since} \ \pi' \ \text{hom})$$ $$\implies$$ $$(\delta \circ \pi)(c) = 0 \iff \delta(\pi(c)) = 0 \implies \pi(c) = 0 \quad(\delta \ \text{being injective}).$$
So $$c \in \operatorname{ker}(\pi) \implies c \in \operatorname{Im}(\varphi)$$ so that $$\exists b \in B\;|\;\varphi(b) = c.$$
Hence we get $$\gamma \circ c = (\gamma \circ \varphi)(b) \underbrace{=}_{\text{comm. diagram}} (\varphi' \circ \beta)(b) = 0$$ $$\implies$$ $$\beta(b) \in \operatorname{ker}(\varphi') \implies \beta(b) \in \operatorname{Im}(\psi')$$ so that $$\exists a' \in A'\;|\; \psi'(a') = \beta(b)$$ we further note that since $\alpha$ is surjective, there exists $$a \in A$$ so that $$(\psi' \circ \alpha)(a) = \psi'(a') = \beta(b)$$ and since the diagram commutes, we have $$(\beta \circ \psi)(a) = (\psi' \circ \alpha)(a) = \psi'(a') = \beta(b)$$ now since $\beta$ is injective, we get $$\beta(b) = \beta(\psi(a)) \implies \psi(a) = b$$ and since $$\operatorname{Im}(\psi) = \operatorname{ker}(\varphi)$$ we have that $$\psi(a) = b \implies \varphi(b) = 0 = c$$ which is what we wanted to show (I believe).
Is this argument correct? If not, where is it wrong?
Thank you in advance.