# Why Homology is homotopy invariant

I've red Hatcher's proof of homotopy invariance for singular homology. I understood how the proof works and the reason why we construct the prism operator (i.e. its a chain contraction). However, this proof gave zero intuition relative to why is homology homotopy invariant. Its clear how to show it, but not at all why is it true. I would appreciate any form of intuitive explanation (geometric, topological..). Thank you in advance.

• I think it gives plenty of intuition: a homotopy between two chains will, through the prism operator, give a homology between them. Commented Mar 22, 2018 at 17:20
• Homotopy invariance of singular homology is easier to prove in a cubical approach, as in Massey's book on "Singular Homology". The reason for this is that if $I^n$ denoted the standard $n$-cube, then $I^{n+1} \cong I^n \times I$, so that homotopies fit better in the cubical framework than in the simplicial one. Commented Mar 23, 2018 at 14:35
• Could you be a little more specific? The proof shows that from an homotopy between spaces you can build an homotopy of chain complexes, and homotopy of chain complexes grants the equality of the maps in homotopy. Which part does troubles you exactly? Commented Apr 19, 2018 at 9:57

This is imprecise, but I think in the following way: The cellular chain complex $C_*(I)$ of the unit interval $I$ is $0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}^2 \rightarrow 0$ concentrated in degrees one and zero. The nonzero differential is $\begin{bmatrix}1 \\ -1\end{bmatrix}$. A continuous map $X \times I$ to $Y$ (in other words a topological homotopy) "should" (this should be justified) induce a chain map from $C_*(X) \otimes C_*(I)$ (tensor product of chain complexes) to $C_*(Y)$. I claim that such a chain map defines a chain homotopy. See below for a proof. My takeaway is that the above chain complex serves the same purpose in the category of chain complexes as the unit interval does in the category of spaces.
To see this, let's consider the tensored chain complex. At each degree $k$, we have \begin{align*} [C_*(X) \otimes C_*(I)]_k &= \left(C_k(X) \otimes \mathbb{Z}^2\right) \oplus \left(C_{k-1}(X) \otimes \mathbb{Z}\right) \\ &\cong C_k(X) \oplus C_k(X) \oplus C_{k-1}(X) \end{align*} The $k$-th differential of this chain complex can be identified with the matrix $\begin{bmatrix}d_k^X & 0 & (-1)^{k-1} \\ 0 & d_k^X & (-1)^k \\0 & 0 & d_{k-1}^X\end{bmatrix}$, where $d_k^X: C_k(X) \rightarrow C_{k-1}(X)$ is the $k$-th differential for $X$.
Finally, if the morphisms \begin{align*} \begin{bmatrix} f_k & g_k & h_k \end{bmatrix} \colon C_k(X) \oplus C_k(X) \oplus C_{k-1}(X) \rightarrow C_{k}(Y) \end{align*} for various $k$ are going to assemble into a chain map, they have to commute with the differentials: \begin{align*} d_k^Y \begin{bmatrix} f_k & g_k & h_k \end{bmatrix} &= \begin{bmatrix} f_{k-1} & g_{k-1} & h_{k-1} \end{bmatrix} \begin{bmatrix}d_k^X & 0 & (-1)^{k-1} \\ 0 & d_k^X & (-1)^k \\0 & 0 & d_{k-1}^X\end{bmatrix}\end{align*} Thus, we get \begin{align*}d_k^Y f_k &= f_{k-1} d_k^X \, , \\ d_k^Y g_k &= g_{k-1}d_k^X \, ,\\ d_k^Y h_k &= (-1)^{k-1}f_{k-1} + (-1)^kg_{k-1} + h_{k-1}d_{k-1}^X \, .\end{align*} We may rewrite the last line as $d_k^Y (-1)^{k-1}h_k + (-1)^{k-2}h_{k-1}d_{k-1}^X = f_{k-1} -g_{k-1}$. We see that the $f_k$'s and $g_k$'s assemble into chain maps $f_*$ and $g_*$, and the $(-1)^{k-1}h_k$'s form a chain homotopy from $f_*$ to $g_*$.