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.
 A: 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.
