# Homologous chain and chain homotopy

Let $$(C,\partial)$$ be a chain complex and $$\varphi,\psi\colon C_\bullet\longrightarrow C_\bullet$$ chain morphisms.

Suppose $$\varphi$$ is homotopic to $$\psi$$, i.e. there exists $$T:C_\bullet\longrightarrow C_{\bullet+1}$$ with $$\partial_{p+1}\circ T_p+T_{p-1}\circ\partial_p=\varphi_p-\psi_p$$.

My question is: if $$c\in C_p$$ then is $$\varphi_p(c)-\psi_p(c)\in B_p(X) ?$$

This is obvious when $$c\in Z_p$$ because $$\varphi_p(c)-\psi_p(c)=\partial_{p+1}\circ T_p(c)$$, but in general?

I would apply this to $$C=S(X)$$ the singular chain complex of a topological space $$X$$ with $$\varphi=Sd$$ (Suddivision operator) and $$\psi=id$$. In fact $$Sd\sim id$$ and I know that if $$A,B\subseteq X$$ are open with $$X=A\cup B$$ then for $$c\in S_p(X)$$ there is a $$k$$ such that $$Sd^k (c)=c_1+c_2$$ ($$k$$ iterated composition) with $$c_1\in S_p(A), c_2\in S_p(B)$$.

I would write $$c=c_1+c_2+b$$ with $$b\in B_p(X)$$ for a proof of excision and Mayer-Vietoris.

This isn't true, choose any $$T$$ so that $$\partial T \partial$$ isn't zero. Then let $$\phi = \partial T + T \partial$$.
$$\phi$$ will be a chain map since $$\partial (\partial T + T \partial) = \partial T \partial = (\partial T + T \partial) \partial$$.
Then $$T$$ is a chain homotopy between $$\phi$$ and the zero map, $$\phi(c) - 0(c)$$ can't always be a boundary since if it was $$\partial(\phi - 0)$$ would have to be zero. But that can't happen since it is equal to $$\partial T \partial$$ which can never be zero by definition.
Here is an explicit construction of such a $$T$$.
Let $$C_*$$ be the chain complex given by $$C_0 = \mathbb Z$$, $$C_1 = \mathbb Z$$ and $$\partial:C_1 \rightarrow C_0$$ be given by the identity, let $$C_i = 0$$ for all other $$i$$. We only have to define $$T$$ on $$C_0$$ since it is trivial on all other degrees. We define $$T:C_0 \rightarrow C_1$$ to just be the identity. Then $$\partial T \partial: C_1 \rightarrow C_0$$ is not zero, it is just the identity morphism $$\mathbb Z \rightarrow \mathbb Z$$ in fact.