# Homotopy and chain homotopy determine each other

In continuation of this previous question, I'm having problems with the following proposition from Kamps and Porter's Abstract Homotopy and Simple Homotopy Theory:

Proposition (3.7). [page 210]

• If $$H:X\otimes I \rightarrow Y$$ is a homotopy between $$f$$ and $$g$$, then defining $$h(z)=H(0,0,z)$$ gives a chain homotopy between $$f$$ and $$g$$.
• If $$h$$ is a chain homotopy between $$f$$ and $$g$$ then $$H(x,y,z)=f(x)+g(y)+h(z)$$ gives a homotopy, $$H:X\otimes I \rightarrow Y$$ between $$f$$ and $$g$$.

The proof is left to the reader. I tried to work it out but failed. Here are the details:

As in the previous question, $$(X\otimes I)_n=X_n \oplus X_n \oplus X_{n-1}$$. First of all, we have the cylinder object $$(X\otimes I,e_0, e_1, \sigma)$$ defined by $$e_0:X\rightarrow X\otimes I,\;x\mapsto (x,0,0)$$, $$e_1:X\rightarrow X\otimes I,\;y\mapsto (0,y,0)$$, and $$\sigma :X\otimes I\rightarrow X,\;(x,y,z)\mapsto x+y$$.

Define $$i(z):X_{n-1}\rightarrow X_n \oplus X_n \oplus X_{n-1}$$ by $$i:z\mapsto (0,0,z)$$. Now $$h_{n-1}=H_n\circ i$$. For the first part we want to prove the relation $$\partial _n ^Y h_{n-1}+h_{n-2}\partial ^X_{n-1}=g_{n-1}-f_{n-1}$$ where $$h_n:X_n \rightarrow Y_{n+1}$$. I tried calculating separate terms. Since $$H$$ is a homotopy $$g-f=He_1-He_0=H(e_1-e_0)$$ so $$(g-f)(x)=H(-x,x,0)$$. Since $$H$$ is in particular an arrow in $$\mathsf{Ch}_\bullet$$, it commutes with boundaries so $$\partial ^Y _n H_n =H_{n-1}\partial ^\prime _n$$ where $$\partial ^\prime$$ is the boundary of $$X\otimes I$$ as in the previous question. Hence $$\partial _n ^Y h_{n-1}=\partial ^Y _n H_n\circ i =H_{n-1}\circ \partial ^\prime _n \circ i$$ which means $$\partial _n ^Y h_{n-1}(z)=H_n(0,0,-\partial ^X z)=-h_{n-2}\partial ^X _{n-1}(z)$$ and the LHS of the chain homotopy relation is identically zero..

For the second part, I need to verify for instance that $$He_0=f$$. By definition $$(He_0)(x)=H(x,0,0)=f(x)+g(0)+h(0)$$ but if $$f$$ and $$h$$ are homomorphisms, wouldn't $$g(0)=h(0)=0$$? In that case I get $$He_0=f$$ without even using the fact $$h$$ is a chain homotopy..

I would like explanations of my mistakes as well as an actual correct proof of the proposition.

Added: Here is a related question.

For part 1: The part where you (between lines) write $H_{n-1} \circ \partial '_n \circ i_n = -h_{n-2}\circ \partial^X_{n-1}$ is the error. Remember that $\partial '(0,0,z)=(-z,z,-\partial^X z)$. Then you should get your homotopy.
For part 2: To begin with you only know that $H$ is a degreewise homomorphism. The property of being a chain homotopy comes into play when you want to show that $H$ is actually a chain homomorphism.