Isomorphic cokernels giving isomorphic kernels

$\require{AMScd}$ This is an issue with a proof related to a characterization of flatness by Eisenbud (in his "Commutative Algebra with a View Toward Algebraic Geometry", p. 162).

If we have a commutative diagram (of $R$-modules)

\begin{CD} @. 0 @. 0 \\ @. @VVV @VVV\\ @. K @. K'\\ @. @VVV @VVV\\ 0 @>>> A @>f>> A'@>f'>> A'' @>>> 0\\ @. @V\alpha VV @V\alpha' VV @| @.\\ @. B @>g>> B'@>g'>> A'' @>>> 0\\ @. @V\beta VV @V\beta' VV @. @.\\ @. C @= C\\ @. @VVV @VVV\\ @. 0 @. 0 \end{CD}

with exact columns and rows (note with cokernels in the columns equal and in the rows equal), then $f$ restricts to an injection $K\to K'$. Indeed if $x\in K$ then $\alpha'(f(x))=g(\alpha(x))=0$ so $f(x)\in K'$. Eisenbud then states in his argument that this map is an isomorphism. Does this follow?

The snake lemma doesn't naturally apply since the second row isn't assumed to begin with a monomorphism. From $\alpha'(y)=0$ we can argue that $y=f(x)$ for some $x$ (since $f'(x)=g'(\alpha'(x))=0$), and then we have $g(\alpha(x))=\alpha'(f(x))=0$, but can we then obtain that $\alpha(x)=0$?

I can't find errata to that book, but it is possible that Eisenbud made a mistake or that the result doesn't follow directly from the diagram.

\begin{CD} @. 0 @. 0 \\ @. @VVV @VVV\\ @. 0 @. R\\ @. @VVV @VVV\\ 0 @>>> R @>=>> R@>0>> 0 @>>> 0\\ @. @V= VV @V0 VV @| @.\\ @. R @>0>> 0@>0>> 0 @>>> 0\\ @. @V0 VV @V0 VV @. @.\\ @. 0 @= 0\\ @. @VVV @VVV\\ @. 0 @. 0 \end{CD}