**Hatcher Exercise 2.1.13**

Here, we verify that two homotopic maps induce the same homomorphisms, , for *reduced* homology groups.
This amounts to showing that and remain chain-homotopic when considering the extended chain complex of reduced homology.
So we show:

**Theorem**: If and are (normally) chain-homotopic, then they are chain-homotopic in the setting of reduced homology.

*Proof*:
If and are (normally) chain-homotopic, we have maps that satisfy for all and also the *terminal condition* .

Consider the extended chain complex, where are defined to be the identity map on the extra . In order to have a chain-homotopy now, we need to define a map in addition to the ’s:

Clearly, for , the the chain homotopy condition is still satisfied by the ’s. We must require in addition that

- (1)
- (2) (the terminal condition)

Observe that the terminal condition actually forces to be the zero map (since is injective). Letting be the zero map, the first condition becomes , which is precisely the original terminal condition that we know satisfies. So, taking to be the zero map satisfies both conditions, and makes the collection of maps into a (reduced) chain-homotopy between and .