Suppose we have a short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ and a projective resolution $P$ for $A$ and a projective resolution for $C$, Q. Is there a way we can construct a projective resolution for $B$ of the form $P \bigoplus Q$ in a way that we can have a short exact sequence of complexes $ 0 \rightarrow P \rightarrow P \bigoplus Q \rightarrow Q \rightarrow 0$ ? By this i mean can we construct specific maps for the resolution in $B$ in order to be a resolution and a chain transformation? Im just looking for an yes or no answer, i have tried to do this on my own but i havent quite managed to construct the specific maps myself, but would like to try some more if the answer is true. Thanks in advance.
Constructing a resolution from other projective resolutions and a short exact sequence
homological-algebrahomology-cohomologyprojective-module
Related Solutions
Claim 1 is indeed wrong : the short exact sequence $0\to\mathbb{Z/2Z}\to\mathbb{Z/4Z}\to\mathbb{Z/2Z}\to 0$ does not define a distinguished triangle in the homotopy category. Indeed by definition a triangle $A\to B\to C\to A[1]$ is distinguished iff it is isomorphic to a triangle $A\to B\to\operatorname{Cone}(A\to B)\to A[1]$, but $C=\operatorname{Cone}(\mathbb{Z/2Z\to Z/4Z})$ is not isomorphic to $\mathbb{Z/2Z}$ is the homotopy category.
To see this, note that if you have an isomorphism $C\simeq \mathbb{Z/2Z}$ it will have to be represented by a morphism of complexes $\mathbb{Z/2Z}\to C$ (this is because $Hom_K=Hom_C/\sim_{htp}$), in other words, there will be a morphism of complexes : $$\require{AMScd} \begin{CD} 0@>>>0@>>>\mathbb{Z/2Z}@>>>0\\ @.@VVV@VVV@.\\ 0@>>>\mathbb{Z/2Z}@>>>\mathbb{Z/4Z}@>>>0 \end{CD}$$
which is an isomorphism in the homotopy category.
This is easy so see that there is only two such morphisms. But none of them are a homotopy equivalence : in fact they are not even a quasi-isomorphism !
This argument fail in the derived category because morphisms $\mathbb{Z/2Z}\to C$ does not necessarily comes from a morphism of complexes. We have in fact $$Hom_D(\mathbb{Z/2Z},C)=Hom_D(P,C)=Hom_K(P,C)=Hom_C(P,C)/\sim_{htp}$$ where $P$ is any projective resolution of $\mathbb{Z/2Z}$.
Take your favorite resolution of $\mathbb{Z/2Z}$ (for example $P=\mathbb{Z\overset{\times 2}\to Z}$), you will see that there is a quasi-isomorphism $P\to C$. Even more : there is also a morphism $P\to\mathbb{Z/2Z}[1]$ leading to a triangle : $$\mathbb{Z/2Z\to Z/4Z}\to P\to\mathbb{Z/2Z}[1]$$ which is isomorphic in the derived category to $\mathbb{Z/2Z\to Z/4Z}\to C\to\mathbb{Z/2Z}[1]$.
Claim 2 and Claim 3 are true but require a precision. Claim 2 follows from Claim 3, so let's have a look at Claim 3.
The correct claim is the following : if you have a short exact sequence of complexes $0\to A\to B\to C\to 0$, then there is a map $C\to A[1]$ in the derived category such that $A\to B\to C\to A[1]$ is a distinguished triangle.
In particular :
- this is not true for any map $C\to A[1]$
- the map $C\to A[1]$ only lives in the derived category and not necessarily at the level of complexes (as you can see by taking a short exact sequence of complexes concentrated in a single degree as above).
From your linked thread (Short exact sequence makes exact triangle in derived category) you know that there is a quasi-isomorphism $\varphi:\operatorname{Cone}(A\to B)\to C$.
So the map $C\to A[1]$ is constructed as : take the inverse of $\varphi$, that is $\varphi^{-1}:C\to\operatorname{Cone}(A\to B)$ and compose with $\operatorname{Cone}(A\to B)\to A[1]$.
(Note however that $\varphi^{-1}$ only exists in the derived category.)
With this construction of $C\to A[1]$, the claim is now obvious since by construction : $$\begin{CD} A@>f>> B@>i>>\operatorname{Cone}(A\to B)@>p>> A[1]\\ @|@|@V\varphi VV@|\\ A@>f>> B@>g>> C@>p\circ \varphi^{-1}>> A[1] \end{CD}$$ commutes and so is an isomorphism of triangles in the derived category.
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.
Best Answer
Yes, this is called the Horseshoe Lemma.