Is the image of a idempotent morphism in $\mathcal{K}(\mathcal{A})$ defined in the naive way?

Question

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 $$ Y\cong \alpha(X) \text{ in } \mathcal{K}(\mathcal{A})? $$

Answer 1

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 $\mathcal{K}(\mathcal{A})$.