## § Homotopic maps produce same singular homology: Intuition

Take two maps $f, g: X \rightarrow Y$ which are homotopic. We wish
to show that if $f$ is homotopic to $g$, then we will get the same
induced singular homology groups from $f$ and $g$. The idea is to take a chain
$c[n] \in C[n]$, then study the image $f\sharp(c[n])$ and $g\sharp(c[n])$. Since $f$
and $g$ are homotopic, we can build a "prism" that connects $f\sharp(c[n])$ and $g\sharp(c[n])$
by means of a prism operator, that sends a chain $c \in C[n]$
to the prism $p(c)$ produced by the chain which lives in $D[n+1]$. Next, we compute the
boundary of this prism, $\partial p(c)$. This boundary will contain a top portion,
which is $f\sharp(c)$, a bottom portion which is $g\sharp(c)$, and the boundary edges
of the prism itself, which is the same as taking the prism of the boundary edges
$p(\partial c)$. This gives the equation $\partial(p(c)) = p(\partial c) + g\sharp(c) - f\sharp(c)$.
Rearranging, this gives $g\sharp(c) - f\sharp(c) = \partial p(c) - p(\partial c)$. To inspect
homology, we wish to check that $f\sharp, g\sharp$ agree on elements of $\ker(\partial[n])/im(\partial[n+1])$.
So, we set $c \in \ker(\partial[n])$. This kills of $p(\partial c)$. Further, since we
quotient by $\partial[n+1]$, the $\partial(p[c])$ also dies off. This means that
$g\sharp(c) - f\sharp(c) = 0$ when interpreted as an element of $H[Y][n]$. Philosophically,
living in $\ker(\partial[n])$ kills $- \circ \partial$, and quotienting
by $im(\partial[n+1])$ kills $\partial \circ -$. Thus, we get that $g\sharp$ and $f\sharp$ 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
$f\sharp$ into $g\sharp$ by raising $f\sharp$ into $g\sharp$ one vertex at a time. This then allows us
to induce cancellations and show that $\partial(p(c))$ contains the terms
$p(\partial(c))$, $f\sharp(c)$ and $g\sharp(c)$.
Let's consider a 1D line $l: \Delta^1 \rightarrow X$ in $X$, with vertices
$l[0], l[1]$. The image of this line under $f$ is $m \equiv f \circ l$ with $m[0], m[1]$ as vertices,
and under $g$ is $n$ with $n[0], n[1]$ as vertices. Let $H: [0, 1] \times [0, 1] \rightarrow X$
be the homotopy between $H(0) f$ and $H(1) = g$. The prism is image of the function $p: [0, 1] \times \Delta^1$, defined
as $p(t, i) \equiv H(t, l(i))$. We see that $p(0, i) = H(0, l(i)) = f(l(i)) = m$, and $p(1, i) = g(l(i)) = n$.
So, we get a "prism" whose endpoints are $m = f \circ l$ and $n = g \circ l$.