Theorem: If $f, g : X \to Y$ are homotopic maps via $H : X \times [0,1] \to Y$ with $H(x, 0) = f(x)$ and $H(x, 1) = g(x)$, then $f_* = g_* : H_n(X) \to H_n(Y)$ for all $n$.

Strategy: Construct a chain homotopy $P : C_n(X) \to C_{n+1}(Y)$ satisfying: $\partial P + P\partial = g_\# - f_\#$ on chain groups. This algebraic relation implies $f_* = g_*$ on homology.

Step 1: Defining the prism operator

For a singular simplex $\sigma : \Delta^n \to X$, define the prism $\sigma \times I : \Delta^n \times [0,1] \to X \times [0,1]$ by $(\sigma \times I)(t, s) = (\sigma(t), s)$.

Compose with $H$ to get $H \circ (\sigma \times I) : \Delta^n \times [0,1] \to Y$. The prism $\Delta^n \times [0,1]$ can be subdivided into $(n+1)$-simplices.

Step 2: Subdividing the prism

Write $\Delta^n \times [0,1]$ as a union of $(n+1)$-simplices. Standard construction: if $\Delta^n$ has vertices $v_0, \ldots, v_n$, then $\Delta^n \times [0,1]$ has vertices $(v_i, 0)$ and $(v_i, 1)$. Subdivide using the $(n+1)!$ simplices corresponding to orderings.

Define $P(\sigma)$ as the signed sum of these $(n+1)$-simplices composed with $H \circ (\sigma \times I)$.

Step 3: Computing boundaries

The boundary of $\Delta^n \times [0,1]$ consists of:

• Top face: $\Delta^n \times \{1\}$ (where $H$ equals $g$)
• Bottom face: $\Delta^n \times \{0\}$ (where $H$ equals $f$)
• Side faces: $(n+1)$-simplices with vertices $(v_i, 0)$ and $(v_i, 1)$ omitted alternately

Computing $\partial P(\sigma)$ using the boundary formula: $\partial P(\sigma) = g(\sigma) - f(\sigma) - P(\partial \sigma)$

Rearranging: $\partial P + P\partial = g_\# - f_\#$ on $C_n(X)$.

Step 4: Descending to homology

Let $[\alpha] \in H_n(X)$, so $\partial \alpha = 0$. Then: $g_\#(\alpha) - f_\#(\alpha) = \partial P(\alpha) + P(\partial \alpha) = \partial P(\alpha)$

Thus $g_\#(\alpha) - f_\#(\alpha) \in B_n(Y)$ is a boundary, so $[g_\#(\alpha)] = [f_\#(\alpha)]$ in $H_n(Y)$.

Therefore $g_*([\alpha]) = f_*([\alpha])$ for all $[\alpha] \in H_n(X)$, proving $f_* = g_*$. ∎