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 $\alpha$ and $\gamma$ are surjective,and $\delta$ is injective.
I want to show that it follows that $\beta$ is surjective.
Here is my reasoning:
Take arbitrary $$b' \in B' \implies \varphi'(b') \in \operatorname{ker}(\pi').$$
Since $\gamma$ is surjective, for each $b' \in B'$, exists $c \in C$ such that $$\gamma(c) = \varphi'(b')$$
For such a $c$ (associated to a $b'$) we have $$(\pi' \circ \gamma)(c) \underbrace{=}_{\text{diagram commutes}} (\delta \circ \pi)(c) = 0$$
$$\delta(\pi(c)) = 0 \implies \pi(c) = 0 \implies c \in \operatorname{ker}(\pi).$$
By exactness, we have $$c \in \operatorname{Im}(\varphi).$$ That is, there exists $b \in B$ such that $$\varphi(b) = c.$$
Hence we have $$\gamma(c) = (\gamma \circ \varphi)(c) = \varphi'(b').$$
By commutativity of the diagram, we have
$$(\gamma \circ \varphi)(b) = (\varphi' \circ \beta)(b) = \varphi'(b')$$ so that, since $\varphi'$ is a homomorphism, we get
$$\varphi'(\beta(b)) = \varphi'(b') \iff \varphi'(\beta(b)-b') = 0$$ $$\implies$$ $$\beta(b)-b' \in \operatorname{ker}(\varphi') \underbrace{\implies}_{\text{exactness}} \beta(b)-b' \in \operatorname{Im}(\psi).$$
It follows that there exists $a' \in A'$ such that $$\psi'(a') = \beta(b)-b'.$$ Because of the surjectivity of $\alpha$, there exists $a \in A$ such that $$\alpha(a) = a'$$ so that $$(\psi' \circ \alpha)(a) = \beta(b)-b'$$ and since the diagram commutes, we have $$(\beta \circ \psi)(a) = \beta(b)-b' \iff b' = \beta(b)-\beta(\psi(a)) \underbrace{\iff}_{\beta \ \text{homomorphism}} b' = \beta(b-\psi(a)).$$
Since $$b-\psi(a) \in B$$ we see that for arbitrary $b' \in B'$, there exists $b-\psi(a) \in B$ so that $$\beta(b-\psi(a)) = b'.$$
Is this proof correct? If not, what is wrong?