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The following content is from the note from MIT.

Let $X,Y$ be topological space. Let $h:f_0 \simeq f_1: X \to Y$ be a homotopy between two continuous maps. The claim we want to prove is that it induces a chain homotopy $f_{0*} \simeq f_{1*}:S_*(X) \to S_*(Y)$, where $S_n(X)$ is the vector space of singular $n$-chains and $f_{0*}:S_*(X) \to S_*(Y)$ is defined by affine extension of the map $\sigma \to f_{0} \circ \sigma$, $\sigma:\Delta_n \to X$ is a simplex.

Let $C_*,D_*,E_*$ be two chain complexes. It can be proved that given a chain homotopy $k:f \simeq g:C_* \to D_*$ and a chain map $j:D_* \to E_*$, then $j \circ k:C_n \to E_{n+1}$ is a chain homotopy between $j\circ f$ and $j \circ g$. (***)

What I am confused is the way the author uses the result.

Let $\iota_0,\iota_1:X \to X \times [0,1]$ be two inclusions. The author would like to construct a chain homotopy between these two maps. Note that these two maps induce two maps between homologies $H_n(\iota_0),H_n(\iota_1):H_n(X) \to H_n(X \times [0,1])$. The author wants to construct a chain homotopy $k:H_n(X) \to H_{n+1}(X \times [0,1])$. Moreover, we also have $H_n(h):H_n(X) \to H_n(Y)$ as a chain map.

If we let $C_n=H_n(X),D_n=H_n(X \times[0,1]),E_n=H_n(Y)$, then the chain map is not quite in the same direction as the one in (***). What is wrong here?

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No, $H_n(h) : H_n(X\times [0,1])\to H_n(Y)$ !

(Moreover, note that you wrote $H_n$ but every time it was supposed to be $S_*$, you haven't passed to homology yet, chain homotopies are built on the chain complex level; so what I should have answered was : "No, $S_n(h) : S_n(X\times [0,1])\to S_n(Y)$")

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  • $\begingroup$ Oh! What a stupid mistake I have made! Thank you! $\endgroup$
    – Jerry
    Jun 27 '19 at 3:15

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