# 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! Posted on Categories Articles