# How to construct a short exact sequence of complexes

Suppose that a hsort exact sequence

$$0 \longrightarrow A \overset{f}{\longrightarrow B} \overset{g}{\longrightarrow} C \longrightarrow 0$$

of objects in some (Abelian) category is given. Also, assume that each may have a resolution

$$0 \longrightarrow A \longrightarrow R_0(A) \longrightarrow R_1(A)\longrightarrow \cdots$$ (similarly for $$B$$ and $$C$$).

I would like to construct a short exact sequence of complexes

$$0 \longrightarrow C^\bullet(A) \longrightarrow C^\bullet(B) \longrightarrow C^\bullet(C) .$$

Probably, the objects $$R_k(A)$$ (and similarly) must satisfy some condition, so we will assume that our category has enough injective and that each $$R_k(A),R_k(B)$$ and $$R_k(C)$$ is injective.

How can I construct my desired exact sequence? I thoguht of taking the pushout of $$f:A\rightarrow B$$ and $$i_0(A):A\rightarrow R_0(A)$$, and then inject it into an injective object I will call $$R_0(B)$$. Then, I could take the cokernel of $$R_0(A)\hookrightarrow R_0(B)$$, but I will not obtain a commutative square ($$C$$ may not inject into it even!).

Another option is to consider the pushout of $$g$$ and $$i_0(B):B\rightarrow R_0(B)$$. It should contain the pushout of $$0:A\rightarrow C$$ and $$i_0(X):A\rightarrow R_0(B)$$ as a subobject, but since I requiere each $$R_k(-)$$ to be injective, I would need to define $$R_0(C)$$ containing $$R_0(B)\coprod_Y C$$, so I would lose surjectivity in case I had it before.

Remark. I think the quiestion could seem a little bit confusion but I cannot restate it to meake it clearer, so let me explain you my goal. A way of seeing the right derived functor of a left-exact functor is as a canonical way of continuing the exact sequence

$$0 \longrightarrow FA \longrightarrow FB \longrightarrow FC .$$

This is easy to see using the Zig-zag lemma once you have the short exact sequence of complexes. Then, my goal is to construct such a sequence from the short exact sequence given above and the injective resolution of each object.

Another method could be to use the Horseshoe lemma. If you have a resolution of $$A$$ and $$C$$, then you can construct a resolution of $$B$$ just by setting $$R_{k}(B)=R_{k}(A)\oplus R_{k}(C)$$. This will then also give you a short exact sequence of complexes $$0\to R(A)\to R(B)\to R(C)\to 0$$, since it is degree-wise exact.

Obviously if you already have a resolution of $$B$$, then it will be homotopic to the resolution obtained by the horseshoe lemma (assuming you're taking resolutions with a sufficiently nice class, like injectives).

Chapter 8 of Relative Homological Algebra by Enochs and Jenda is a good place for this kind of discussion.

Edit. The desired map can be constructed as follows. Let $$\alpha:A\to R_{0}(A)$$ and $$\gamma:C\to R_{0}(C)$$ be the embeddings at the start of the injective resolutions, and let $$f:A\to B$$ and $$g:B\to C$$ be the maps in the short exact sequence. Then by injectivity of $$R_{0}(A)$$ there is a map $$\epsilon:B\to R_{0}(A)$$ such that $$\alpha = \varepsilon\circ f$$. Define a map $$B\to R_{0}(A)\oplus R_{0}(C)$$ via $$b\mapsto (\varepsilon(b),\gamma g(b))$$. This map is injective because if $$b\mapsto (0,0)$$, then as $$\gamma$$ is injective we see that $$b\in\text{ker}(g)=\text{im}(f)$$ so $$b=f(a)$$. Then $$0=\varepsilon(b)=\varepsilon f(a)= \alpha(a)=0$$ so $$a=0$$ as $$\alpha$$ is injective.

For an element free proof, there is still a map $$B\to R_{0}(A)\oplus R_{0}(C)$$ for the reasons as above. By considering the snake lemma, you can quickly see that this is injective.

• Could you provide some details about how $B$ injects into $R_0(A)\oplus R_0(C)$? I would like to have some details about the maps to be able to check that the digram actually commutes. I will also hava a look at the book you mention. Thanks. Feb 18, 2020 at 17:44
• It is a good explanation, but assumes that the category is a concrete category. I had problems showing the injectivity of the map $b\mapsto (\epsilon(b),\gamma g(b))$ when no elements can be considered. Your proof is right if every Abelian category were a concrete category but I think that is not true in general, is it? Feb 18, 2020 at 19:38