3
$\begingroup$

Let $\mathcal{A}$ be an abelian category and $\mathcal{K}(\mathcal{A})$ be the homotopy category of chain complexes in $\mathcal{A}$. It is well-known that $\mathcal{K}(\mathcal{A})$ is idempotent complete in the following sense: Let $X$ be an object in $\mathcal{K}(\mathcal{A})$ and $\alpha: X\rightarrow X$ be a morphism which satisfies $$ \alpha^2=\alpha \text{ in } \mathcal{K}(\mathcal{A}), $$ then there exist $Y$ in $\mathcal{K}(\mathcal{A})$ together with $i: Y\rightarrow X$ and $p: X\rightarrow Y$ such that $$ pi=id_Y \text{ and } ip=\alpha. $$

Now since $X$ is a chain complex and $\alpha$ is a chain map, we can form the image of $\alpha$, $\alpha(X)$ which is a subcomplex of $X$. $\textbf{My question}$ is: do we have an isomorphism $$ Y\cong \alpha(X) \text{ in } \mathcal{K}(\mathcal{A})? $$

$\endgroup$

1 Answer 1

6
$\begingroup$

No. For example, let $A$ be any object of $\mathcal{A}$, let $X$ be the complex $$\dots\to0\to A\oplus A\stackrel{\begin{pmatrix}1&0\end{pmatrix}}{\longrightarrow}A\to0\to\dots$$ and $\alpha:X\to X$ the chain map that is $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ on $A\oplus A$ and zero on $A$. Then $\alpha$ is null-homotopic and so in particular $\alpha^2=\alpha$ in $\mathcal{K}(\mathcal{A})$ and the object $Y$ in the question must be contractible. But $\alpha(X)$ is $$\dots\to0\to A\to0\to0\to\dots,$$ which isn't contractible if $A\not\cong 0$.

In fact, if you allow yourself to replace the object $X$ in the question by some object $X'$ with an isomorphism $X\to X'$ in $\mathcal{K}(\mathcal{A})$ and $\alpha$ by some map $\alpha':X'\to X'$ that makes the obvious square commute in $\mathcal{K}(\mathcal{A})$, there is no restriction at all on the isomorphism class of $\alpha'(X')$ in $\mathcal{K}(\mathcal{A})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.