§ Homotopic maps produce same singular homology: Intuition

Take two maps f,g:XYf, g: X \rightarrow Y which are homotopic. We wish to show that if ff is homotopic to gg, then we will get the same induced singular homology groups from ff and gg. The idea is to take a chain c[n]C[n]c[n] \in C[n], then study the image f(c[n])f\sharp(c[n]) and g(c[n])g\sharp(c[n]). Since ff and gg are homotopic, we can build a "prism" that connects f(c[n])f\sharp(c[n]) and g(c[n])g\sharp(c[n]) by means of a prism operator, that sends a chain cC[n]c \in C[n] to the prism p(c)p(c) produced by the chain which lives in D[n+1]D[n+1]. Next, we compute the boundary of this prism, p(c)\partial p(c). This boundary will contain a top portion, which is f(c)f\sharp(c), a bottom portion which is g(c)g\sharp(c), and the boundary edges of the prism itself, which is the same as taking the prism of the boundary edges p(c)p(\partial c). This gives the equation (p(c))=p(c)+g(c)f(c)\partial(p(c)) = p(\partial c) + g\sharp(c) - f\sharp(c). Rearranging, this gives g(c)f(c)=p(c)p(c)g\sharp(c) - f\sharp(c) = \partial p(c) - p(\partial c). To inspect homology, we wish to check that f,gf\sharp, g\sharp agree on elements of ker([n])/im([n+1])\ker(\partial[n])/im(\partial[n+1]). So, we set cker([n])c \in \ker(\partial[n]). This kills of p(c)p(\partial c). Further, since we quotient by [n+1]\partial[n+1], the (p[c])\partial(p[c]) also dies off. This means that g(c)f(c)=0g\sharp(c) - f\sharp(c) = 0 when interpreted as an element of H[Y][n]H[Y][n]. Philosophically, living in ker([n])\ker(\partial[n]) kills - \circ \partial, and quotienting by im([n+1])im(\partial[n+1]) kills \partial \circ -. Thus, we get that gg\sharp and ff\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 ff\sharp into gg\sharp by raising ff\sharp into gg\sharp one vertex at a time. This then allows us to induce cancellations and show that (p(c))\partial(p(c)) contains the terms p((c))p(\partial(c)), f(c)f\sharp(c) and g(c)g\sharp(c). Let's consider a 1D line l:Δ1Xl: \Delta^1 \rightarrow X in XX, with vertices l[0],l[1]l[0], l[1]. The image of this line under ff is mflm \equiv f \circ l with m[0],m[1]m[0], m[1] as vertices, and under gg is nn with n[0],n[1]n[0], n[1] as vertices. Let H:[0,1]×[0,1]XH: [0, 1] \times [0, 1] \rightarrow X be the homotopy between H(0)fH(0) f and H(1)=gH(1) = g. The prism is image of the function p:[0,1]×Δ1p: [0, 1] \times \Delta^1, defined as p(t,i)H(t,l(i))p(t, i) \equiv H(t, l(i)). We see that p(0,i)=H(0,l(i))=f(l(i))=mp(0, i) = H(0, l(i)) = f(l(i)) = m, and p(1,i)=g(l(i))=np(1, i) = g(l(i)) = n. So, we get a "prism" whose endpoints are m=flm = f \circ l and n=gln = g \circ l.