Are both $g', g$ assumed to be surjective in the commutative diagram for $3 \times 3$ lemma?

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The following is taken from Module Theory An Approach to Linear Algebra} by T.S. Blyth

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Theorem 3.4:

Consider the diagram

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of $R$-modules and $R$-morphisms in which $f$ is an $R$-epimorphism. The following conditions are equivalent:

$(1)$ there is a unique $R$-morphism $h:B\to C$ such that $h\cdot f=g$;

$(2)$ $\text{ker }f\subset \text{ker }g$.

Moreover, such an $R$-morphism $h$ is an monomorphism if and only if $\ker f= \ker g$.

Exercise 3.10: [The 3x3 lemma] Consider the following diagram of $R$-modules and $R$-morphisms $\require{AMScd}$ \begin{CD} @. 0 @. 0 @. 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> A' @>{\smash{\alpha'}}>> A @>{\smash{\beta'}}>> A'' @>>> 0 \\ @. @VV{f'}V @VV{f}V @VV{f''}V \\ 0 @>>> B' @>{\smash{\alpha}}>> B @>{\smash{\beta}}>> B'' @>>> 0 \\ @. @VV{g'}V @VV{g}V @VV{g''}V \\ 0 @>>> C' @. C @. C'' @>>> 0 \\ @. @VVV @VVV @VVV \\ @. 0 @. 0 @. 0\end{CD}

Given that this diagram is commutative, that all three columns are exact and that the top two rows are exact, prove that there exist unique $R$-morphisms $\alpha'': C'\to C$ and $\beta'': C\to C''$ such that the resulting bottom row is exact and the complete diagram is commutative.

[Hint. Observe that $g\circ \alpha \circ f' = 0$ so that $\ker g' = \operatorname{im}f' \subset \ker g \circ \alpha$. Use Theorem 3.4 to produce $\alpha''$. Argue similarly to produce $\beta''$. Now chase!]

$\color{Red}{Questions:}$

In the hint for the above exercise (3.10), it cites theorem 3.4 for producing $\alpha'',\beta'',$ after stating that $\ker g' = \operatorname{im}f' \subset \ker g \circ \alpha.$ But the equivalent of $\ker g'$ in theorem 3.4 is $\ker f$, and $f$ in theorem 3.4 is assume to be a $R$-epimorphism and hence surjective. This in turn would make $g'$ to be surjective for the hint to exercise 3.4. Similarly for $g$ in the case of $\beta''$. Since $g',g$ have to assume to be surjective even though it was not made explicit in the hypothesis of the exercsie, can I assume it to be so when doing the exercise. I am not sure why these two minor facts are not made explicit?

Thank you in advance.