homological algebra – Cone of a morphism of complexes that are concentrated in degree $0$ and $1$


Let $R$ be a ring and $f:Ato A’$ and $g:Bto B’$ be morphisms of $R$-modules. Let $h:C_{bullet}to C_{bullet}’$ be a morphism of $R$-module complexes fitting in a morphism of distinguished triangles:

$$require{AMScd}
begin{CD}
C_{bullet} @>>> A(0) @>>> B(0) @>>> C_{bullet}(1) \
@VVhV @VVfV @VVgV @VVh(1)V \
C_{bullet}’ @>>> A'(0) @>>> B'(0) @>>> C_{bullet}'(1)
end{CD}$$

From a paper I am reading, the following is used:

If $f$ and $g$ are injective morphisms, then we have a quasi-isomorphism $operatorname{cone}(h)cong(operatorname{coker}(f)to operatorname{coker}(g))$.

It seems that the equivalent general statement is wrong, i.e. if $f$ and $g$ are not assumed to be injective and $A(0)$, $A'(0)$, $B(0)$ and $B'(0)$ are not assumed to be concentrated in degree $0$ anymore (see for instance this question).

Does anyone have a proof – if true – of this statement?

Many thanks!