# Confusion with a step in the construction of chain homotopy from a homotopy

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?

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, $$S_n(h) : S_n(X\times [0,1])\to S_n(Y)$$")