Homotopy invariance of homology

Remarks

The proof cuts the prism $\Delta^n \times [0, 1]$ into $(n+1)$-simplices. A homotopy $H \colon X \times [0,1] \to Y$ from $f$ to $g$ then gives a prism operator $P \colon C_n(X) \to C_{n+1}(Y)$ with $\partial P + P \partial = g_\# - f_\#$. So $f_\#$ and $g_\#$ are chain homotopic, and chain homotopic maps induce the same map on homology. The second claim follows by applying the first to $g \circ f \simeq \id_X$ and $f \circ g \simeq \id_Y$.

Example: a contractible space $X$, such as $\R^n$ or a convex set, is homotopy equivalent to a point. Hence $H_0(X) \cong \Z$ and $H_n(X) = 0$ for $n \ge 1$. The same holds with any coefficients and for relative homology of pairs, with homotopies of pairs.

Going deeper

For a student

This is the homotopy axiom of the Eilenberg–Steenrod axioms for singular homology. It lets one replace a space by a simpler homotopy equivalent one, for example a deformation retract, before computing. It also shows that singular homology factors through the homotopy category of spaces. Together with excision, the long exact sequence of a pair and the homology of a point, it is enough to compute the homology of spheres and of CW complexes.