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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Dog_69
    Feb 18, 2020 at 17:44
  • $\begingroup$ I'll add this as an edit to my answer $\endgroup$
    – Zeek
    Feb 18, 2020 at 18:14
  • $\begingroup$ 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? $\endgroup$
    – Dog_69
    Feb 18, 2020 at 19:38
  • $\begingroup$ I've added a further edit. $\endgroup$
    – Zeek
    Feb 18, 2020 at 22:37
  • $\begingroup$ I see, thank you. $\endgroup$
    – Dog_69
    Feb 20, 2020 at 13:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .