Homotopy category, map without kernel

Question

I was looking for some examples of an abelian category such that the associated homotopy category is (additive but) not abelian, and I have read an answer to this question. Consider the category of cochain complexes of abelian groups and the map $$\require{AMScd} \begin{CD} \cdots @>>> 0 & @>>> \mathbb{Z} @>>> 0 @>>> \cdots\\ @. @VVV @V{\pi}VV @VVV \\ \cdots @>>> 0 @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 @>>> \cdots \end{CD}$$ We would like to prove that in the homotopy category we don’t have a kernel. The author to the answer considers this map $g^\bullet$ $$\begin{CD} \cdots @>>> 0 & @>>> \mathbb{Z} @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 @>>> \cdots\\ @. @VVV @V{id}VV @VVV \\ \cdots @>>> 0 @>>> \mathbb{Z} @>>> 0 @>>> \cdots \end{CD}$$ Clearly the composition with the given map is homotopic to $0,$ hence there exists a (unique) map $\Phi^\bullet$ from $$\begin{CD} \cdots @>>> 0 & @>>> \mathbb{Z} @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 @>>> \cdots\\ \end{CD}$$ to the kernel $K^\bullet$ such that is $\imath^\bullet\circ \Phi^\bullet$ is homotopic to $g^\bullet$, where $\imath$ is the map that defines the kernel. He claims that no such map can exist, but it’s not clear to me why this is true. Can you give me an explanation?