§ Homotopic maps produce same singular homology: Intuition
Take two maps which are homotopic. We wish
to show that if is homotopic to , then we will get the same
induced singular homology groups from and . The idea is to take a chain
, then study the image and . Since
and are homotopic, we can build a "prism" that connects and
by means of a prism operator, that sends a chain
to the prism produced by the chain which lives in . Next, we compute the
boundary of this prism, . This boundary will contain a top portion,
which is , a bottom portion which is , and the boundary edges
of the prism itself, which is the same as taking the prism of the boundary edges
. This gives the equation .
Rearranging, this gives . To inspect
homology, we wish to check that agree on elements of .
So, we set . This kills of . Further, since we
quotient by , the also dies off. This means that
when interpreted as an element of . Philosophically,
living in kills , and quotienting
by kills . Thus, we get that and produce the same
homology element. Intuitively, we are saying that these can be connected by a prism in the space,
and thus produce the same element. Think of the 0D case in a path-connected
space, where all points become equivalent since we can connect them by paths.
To compute the boundary of the prism, we break the prism into an interplation from
into by raising into one vertex at a time. This then allows us
to induce cancellations and show that contains the terms
, and .
Let's consider a 1D line in , with vertices
. The image of this line under is with as vertices,
and under is with as vertices. Let
be the homotopy between and . The prism is image of the function , defined
as . We see that , and .
So, we get a "prism" whose endpoints are and .