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A homework problem states:

Show explicitly that the chain complex $C_n(X)$ is chain homotopy equivalent to the chain complex: $\dotsm \rightarrow 0 \rightarrow 0 \rightarrow 0 \rightarrow \mathbb{Z}$, where $X$ is contractible.

My attempt:

I started by showing the the given chain complex is homotopic to $\dotsm \overset{id}\rightarrow \mathbb{Z} \overset{0}\rightarrow \mathbb{Z} \overset{id}\rightarrow \mathbb{Z} \overset{0}{\rightarrow} \mathbb{Z}$. I think this may be useful because this is just the chain complex $C_n(Y)$ where $Y$ is a single point.

So I'm taking a null-homotopy $H : X \times I \rightarrow X$ taking the identity to the constant mapping at $p \in X$. I then take $f : X \rightarrow \{p\}$ and $g : \{p\} \rightarrow X$ defined by $g(p)=p$. I'd like to show that $C(f)$ and $C(g)$ give the desired chain homotopy.

For this purpose, I want to define $h_n : C_n(X) \rightarrow C_{n+1}(X)$. I'd like to use the null-homotopy $H$ to let $h_n$ turn a generator of $C_n(X)$ into an element of $C_{n+1}(X)$, but it just doesn't seem to work (something like: $h_n(\varphi)(x_1,\dotsc,x_{n+1},t)=H(\varphi(x_1,\dotsc,x_{n+1}), t)$). This definition does not seem to give the desired chain homotopy.

Note: This is supposed to get us warmed up for a more general theorem.

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Why is there a "$t$" in the expression $h_n(\phi)(x_1, \ldots, x_{n+1}, t)$? If $h_n(\phi) \in C_{n+1}(X)$, then certainly $h_n(\phi)$ is just another simplex. – Shaun Ault Jan 18 '12 at 19:49

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