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I'm trying to understand this one step in the proof of Theorem 1.18.18 (i), the statement of the theorem is as follows:

Let $\mathscr{C}$ be an abelian category and let $$0 \longrightarrow X \overset{f}{\longrightarrow} Y \overset{g}{\longrightarrow} Z \longrightarrow 0$$ be a degreewise split short exact sequence of chain complex over $\mathscr{C}$. The chain map $f$ is a homotopy equivalence if and only if $Z \simeq 0$ (homotopy equivalent).

The place where I'm stuck is the implication: $f$ is a homotopy equivalence implies $Z \simeq 0$. The argument essentially shows that $f$ is in fact a split monomorphism as a chain map and therefore $f$ and $g$ induce an isomorphism $Y \simeq X \oplus Z$. The book then concludes: It follows that $Z \simeq 0$ as $f$ is a homotopy equivalence.

It is this conclusion that I don't follow. Any pointers will be helpful!

I can provide more information if needed, this is on page 127 in the book.

Naweed G. Seldon
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1 Answers1

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Identify $Y$ with $X\oplus Z$, and let $f':X\oplus Z\to X$ be the projection map and $g':Z\to X\oplus Z$ the inclusion map.

In the chain homotopy category $K(\mathscr{C})$, $f$ is an isomorphism, and $f'$ is inverse to $f$. Since $f'g'$ is zero, so is $ff'g'$, which is homotopic to $g'$. Therefore $gg'=\text{id}_Z$ is homotopic to zero. So $Z$ is contractible (homotopy equivalent to the zero complex).

Jeremy Rickard
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