Massey Exercise 2.7.4

Theorem: Let and be the maps and for some fixed . The mapping by is an isomorphism.

Proof: Consider the map defined in Theorem 7.1 – that is, by , where and are the projection maps. This map is an isomorphism. We will show that is the identity, so that . Let and . Then,

Now, note that , and similarly for , so that the product path is:

If we apply , the second half of the path will effectively be nullified – that is:

Clearly, this is a path in that traverses twice as fast, and as such, it’s clearly homotopic to . An explicit homotopy is:

A very similar argument proceeds for the other path in the ordered pair, with the first half of the path being nullified and using the ‘reverse’ homotopy. We then obtain

So, the components of are in the equivalence class of and , respectively, so is the identity on . Then, it must be that , and is itself an isomorphism.

It follows that the elements and commute. It’s easy to see that , and also . So it must be that .