# Chain Homotopy for the Barycentric Subdivision of General Chains

On page 123 Hatcher defines the chain homotopy map $T$ for general chains by
$T(\sigma) = \sigma_{\#}(T(\Delta^n))$. I am having some trouble understanding this. How is $T(\Delta^n)$ a linear combination of singular simplexes of $\Delta^n$? I thought $\sigma_\#$ was a map from $C_{k}(\Delta^n)$ to $C_{k}(X)$?

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In that part Hatcher uses $\Delta^n$ to denote the mapping $1_{\Delta^n} \colon \Delta^n \to \Delta^n$, which is a $n$-chain in the $n$-th module of the singular complex $C_n(X)$.
Your interpretation of $\sigma_\#$ is correct, it's the mapping $$\sigma_\# \colon C_k(\Delta^n) \to C_k(X)$$ which sends every singular simplex of $\Delta^n$ (let's say $\tau \colon \Delta^k \to \Delta^n$) in the singular simplex $\sigma_\#(\tau)= \sigma \circ \tau$.
Now to defined the chain homotopy $T \colon C_k(X) \to C_{k+1}(X)$ Hatcher uses the chain homotopy $T \colon C_k(\Delta^n) \to C_{k+1}(\Delta^n)$, that he defined in the previous paragraph. It defines $T(\sigma)$ (for every $\sigma \colon \Delta^n \to X$) as the image of $\sigma_\# \circ T(\Delta^n)$, where the first $T$ is the chain operator in the complex $C(X)$ while the latter is the chain operator on the complex $C(\Delta^n)$: hence $T(\Delta^n) \in C_{n+1}(\Delta^n)$ and so $\sigma_\#(T(\Delta^n)) \in C_{n+1}(X)$.
Here is what I don't understand. In what sense is $(T(\Delta^n))$ an element of the domain of the induced map? – Jonathan Aronson Sep 18 '13 at 18:27
Do we not need a map from $\Delta^n$ X I to $\Delta^n$? – Jonathan Aronson Sep 18 '13 at 18:38