Here, we verify that two homotopic maps $f \cong g : X \rightarrow Y$ induce the same homomorphisms, $f_* = g_* : \wt{H}_n(X) \rightarrow \wt{H}_n(Y)$, for reduced homology groups. This amounts to showing that $f_\sharp$ and $g_\sharp$ remain chain-homotopic when considering the extended chain complex of reduced homology. So we show:

Theorem: If $f_\sharp$ and $g_\sharp$ are (normally) chain-homotopic, then they are chain-homotopic in the setting of reduced homology.

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Proof: If $f_\sharp$ and $g_\sharp$ are (normally) chain-homotopic, we have maps $P_n : C_n(X) \rightarrow C_{n+1}(Y)$ that satisfy $\delta_{n+2} P_{n+1} + P_n \delta_{n+1} = g_\sharp - f_\sharp$ for all $n \geq 0$ and also the terminal condition $\delta_1 P_0 = f_\sharp - g_\sharp$.

Consider the extended chain complex, where $f_\sharp, g_\sharp : \mathbb{Z} \rightarrow \mathbb{Z}$ are defined to be the identity map on the extra $\mathbb{Z}$. In order to have a chain-homotopy now, we need to define a map $Q$ in addition to the $P_n$'s:

Clearly, for $n > 0$, the the chain homotopy condition is still satisfied by the $P_n$'s. We must require in addition that

• (1) $\delta_1 P_0 + Q \epsilon = f_\sharp - g_\sharp$
• (2) $\epsilon Q = f_\sharp - g_\sharp = 0$ (the terminal condition)

Observe that the terminal condition actually forces $Q$ to be the zero map (since $\epsilon$ is injective). Letting $Q$ be the zero map, the first condition becomes $\delta_1 P_0 = f_\sharp - g_\sharp$, which is precisely the original terminal condition that we know $P_0$ satisfies. So, taking $Q$ to be the zero map satisfies both conditions, and makes the collection of maps $\set{P_n} \cup Q$ into a (reduced) chain-homotopy between $f_\sharp$ and $g_\sharp$.