In an additive category, why a morphism between complexes can be factored as a composition of a homotopy equivalence and a monomorphism?

Question

I was stuck in the following

Problem. Suppose that $\mathcal A$ is an additive category and $f\colon X\to Y$ is a morphism of (cochain) complexes in $\mathcal A$. Then there is a factorization $f=gf'$ where $g$ is a homotopy equivalence and each $(f')^i$ is a monomorphism admitting a retraction.

What I have tried:

• Firstly I considered the mapping cone $\mathrm{Cone}(f)$. Although tried, I was not able to figure out a way of defining morphisms $f'\colon X\to\mathrm{Cone}(f)$ and $g\colon \mathrm{Cone}(f)\to Y$ satisfying $f=gf'$.

• On a second thought, I considered the product $X\oplus Y$. If I define $f'=\left(\begin{smallmatrix}\mathrm{id}_X\\f\end{smallmatrix}\right)\colon X\to X\oplus Y$ and $g=(0,\mathrm{id}_Y)\colon X\oplus Y\to Y$, then $f'$ and $g$ are morphisms between complexes and $f=gf'$. However, though such an $f'$ is a monomorphism and has a retraction, it is not clear to me whether $g$ is a homotopy equivalence.

Hence I would like to ask for some hints for this problem (maybe I lack some prerequisite knowledge and it is not wise to directly construct such a factorization?)

Any help is greatly appreciated.

Answer 1

$\operatorname{Cone}(f)$ won't work since it is a complex which measures the "difference" between $X$ and $Y$ so this is neither equivalent to $X$ nor to $Y$.

Similarly, $X\oplus Y$ won't work since it is a complex which measure the "sum" of $X$ and $Y$, so this is neither equivalent to $X$ nor to $Y$.

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But the Cone idea is a good one, you just need to change it a bit :

Consider $Z=\operatorname{Cone}(f)[-1]$ and the map $u:Z\to X$. Then $\operatorname{Cone}(u)$ does what you want :

• informally, it measures the difference between $X$ and $Z$, and $Z$ is already the difference between $X$ and $Y$, so this is "X-(X-Y)=Y".
• this is not only informal : there is a natural map $\operatorname{Cone}(u)\to Y$ which is a homotopy equivalence (see below).
• there is also a natural map $X\to\operatorname{Cone}(u)$ which is in every degree the inclusion $X^i\to\operatorname{Cone}(u)^i=X^i\oplus Z^{i+1}$ (hence a split injection)

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Now we construct the natural map $P:\operatorname{Cone}(u)\to Y$. First note that $$\operatorname{Cone}(u)^i= X^i\oplus Z^{i+1}=X^i\oplus \operatorname{Cone}(f)[-1]^{i+1}=X^i\oplus Y^i\oplus X^{i+1}$$ and the differential is given by the matrix $$D=\begin{pmatrix} d&0&1\\0&d&f\\0&0&-d\end{pmatrix}$$

• In each degree, define $P:\operatorname{Cone}(u)^i\to Y^i$ by $$\operatorname{Cone}(u)^i=X^i\oplus Y^i\oplus X^{i+1}\xrightarrow{\begin{pmatrix}-f&1&0\end{pmatrix}} Y^i $$
• Check that this map is compatible with the differentials (in other words $PD=dP$), so this is indeed a map $P:\operatorname{Cone}(u)\to Y$.
• We also have a map $I:Y^i\to\operatorname{Cone}(u)^i$ which is given by the natural inclusions : $$Y^i\xrightarrow{\begin{pmatrix}0\\1\\0\end{pmatrix}} X^i\oplus Y^i\oplus X^{i+1}$$ which is also compatible with the differentials ($Id=DI$). So we also have a map $Y\to\operatorname{Cone}(u)$.
• The composite $PI:Y\to\operatorname{Cone}(u)\to Y$ is obviously the identity.
• The composite $IP:\operatorname{Cone}(u)\to Y\to\operatorname{Cone}(u)$ is not the identity, but we can construct an homotopy between the identity and $IP$. This will be the only reasonable map :
  $$ S:X^i\oplus Y^i\oplus X^{i+1}\xrightarrow{\begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix}}X^{i-1}\oplus Y^{i-1}\oplus X^i$$ Check that $SD+DS=\operatorname{id}-IP$

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By the way, this construction is called the mapping cylinder of $f:X\to Y$. If you know some algebraic topology, you can get the intuition from there.