2

Suppose I have a collection of maps defined as follows:

for $d_{n}:C_{n} \rightarrow C_{n-1}$ and $s_{n}: C_{n} \rightarrow C_{n+1}$ I have :

$t_{n}=1-f'_{n} -f_{n}$ , where $f_{n}=s_{n-1}d_{n}$ and $f'_{n}=d_{n+1}s_{n}$.

Furthermore I am given that $s_{n}$ is a collection of maps which satisfies $s_{n+1}s_{n}=0$.

I have already showed that $t^{2}_{n}=t_{n}$.

  • How can I show that such map is chain homotopic to the identity map?
  • If this is the case, does that imply that its image is then itself, since it is in some sense an identity mapping?
feynhat
  • 1,904
Morettin
  • 315

1 Answers1

0

$\text{id} - t_n = f_n + f_n' = s_{n-1}d_n + d_{n+1}s_n$.

Therefore $s_\#$ is the required chain homotopy between $t_n$ and $\text{id}$.

The answer your second question is, no (in general). For example, let $X$ be a topological space such that $X = \text{Int}A \cup \text{Int} B$, let $\iota$ be the inclusion $C_n(A + B) \hookrightarrow C_n(X)$, where $C_n(A+B)$ is the chains group consisting of singular simplices with images either entirely in $A$ or entirely in $B$ and $C_n(X)$ is the singular chain group of $X$.

Then $\iota$ is a chain homotopy equivalence (this is called the barycentric subdivision lemma), with a chain homotopic inverse $\rho : C_n(X) \to C_n(A+B)$.

Note that $\iota \circ \rho : C_n(X) \to C_n(X)$ is homotopic to identity. The image of $\iota \circ \rho$ is obviously not $C_n(X)$, unless $A = B = X$.

What is true however, is that $t_n$ induces identity map at the level of homology $H_n(C_*) \to H_n(C_*)$. So, the image image will be $H_n(C_*)$. This is because if two maps are chain homotopic then the maps induced by them at the homology level are equal.

feynhat
  • 1,904