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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy