Proof of homotopy invariance of homology

Suppose that $f, g : X \to Y$ are homotopic maps, via a homotopy $F$ with $F(x,1) = g(x)$, $F(x,0) = f(x)$. To show that $f_\# , g_\#$ are chain homotopic (and hence $f_* = g_*$ in homology), we first look at dimension $0$. Since these maps are linear, it suffices to look at a singular $0$-simplex $\sigma$; the induced maps are $f_\# (\sigma ) = f\sigma$ and $g_\# (\sigma ) = g\sigma$. Suppose $\sigma (0) = p$. Let $\gamma : [0,1] = \Delta_1 \to Y$ be the singular $1$-simplex defined by the composition $[0,1] \to \left\{ p\right\} \times [0,1] \xrightarrow {F} Y$.

Define $H_0(\sigma ) = \gamma$ and show that $H_0$ provides a chain homotopy between $f\sigma$ and $g\sigma$.

We now look at dimension $1$, needing to define $H_1 : S_1(X) \to S_2(Y)$ so that, for a singular $1$-simplex $\sigma : I \to X$, we have the formula $g_\# \sigma - f_\# \sigma = \partial_2 H_1(\sigma ) + H_0 \partial_1(\sigma )$ (with $H_0$ as in Exercise 6.16.1).

We first use the homotopy $F$ to define a map $K = F(\sigma \times \mathrm{id}) : I \times I \to Y$, where $\sigma \times \mathrm{id} : I \times I \to X \times I$. Note that $Ki_0 = f\sigma$, $Ki_1 = g\sigma$, where $i_0(s) = (s,0)$, $i_1(s) = (s,1)$. If $\tau_1(0) = \sigma (1)$, $\tau_0(0) = \sigma (0)$ are the singular $0$-simplices coming from restricting $\sigma$ to its end points, then $\partial_1 \sigma = \tau_1 - \tau_0$, and $H_0(\tau_1)(y) = K(1,y)$, $H_0(\tau_0)(y) = K(0,y)$.

Constructing H_1.

Motivated by Figure 6.6, define two singular $2$-simplices $\alpha , \beta$ in $I \times I = \Delta_1 \times I$ with positive orientation on $\alpha$ and negative orientation on $\beta$ as depicted via the arrows on their boundaries, and define $H_1(\sigma ) = K\beta - K\alpha = K_\# (\beta - \alpha )$.

Show that with this definition of $H_1$, we have the required formula $g_\# (\sigma ) - f_\# (\sigma ) = \partial_2 H_1(\sigma ) + H_0 \partial_1(\sigma )$.

Use Exercises 6.16.1 and 6.16.2 to show that if $f, g : X \to Y$ are homotopic maps, then the induced maps $f_\# , g_\#$ are chain homotopic in dimensions $0$ and $1$. Deduce that $f_* = g_*$ in homology.

To extend $H_0, H_1$ to higher-dimensional chains, note that the homotopy $F$ is used only at the last step and can be separated from the definition: writing $i_j^X : X \to X \times I$ for $i_j^X(x) = (x,j)$, we have $g = F i_1^X$, $f = F i_0^X$. If we label the maps $H_k = H_k^{X,Y} : S_k(X) \to S_{k+1}(Y)$ already defined ($k=0,1$, Exercises 6.16.1-6.16.2), there are corresponding maps $H_k^X : S_k(X) \to S_{k+1}(X \times I)$ (using the identity homotopy $I(x,t) = (x,t)$ connecting $i_0^X$ and $i_1^X$) so that $H_k^{X,Y}$ factors as $F_\# H_k^X$. The term $(\partial_2 H_1^{X,Y} + H_0^{X,Y}\partial_1)(\sigma )$ becomes $F_\# (\partial H_1^X + H_0^X \partial_1)(\sigma )$, and since $g_\# = F_\# (i_1^X)_\#$, $f_\# = F_\# (i_0^X)_\#$, we can rewrite $g_\# (\sigma ) - f_\# (\sigma ) = F_\# ((i_1^X)_\# (\sigma ) - (i_0^X)_\# (\sigma ))$. So the formula follows from the special case

by composing with $F_\#$.

Let $i_j^{\Delta_1}$ be the corresponding inclusions for $\Delta_1$. Using the identity map $\iota_1 \in S_1(\Delta_1)$, there is a formula $(\partial_2 H_1^{\Delta_1} + H_0^{\Delta_1} \partial_1)(\iota_1) = ((i_1^X)_\# - (i_0^X)_\# )(\iota_1)$; applying $(\sigma \times \mathrm{id})_\#$ to both sides, using that $H_1^X(\sigma ) = (\sigma \times \mathrm{id})_\# (\beta - \alpha ) = (\sigma \times \mathrm{id})_\# H_1^{\Delta_1}(\iota_1)$ (Exercise 6.16.2) and that $(\sigma \times \mathrm{id})_\#$ is a chain map commuting with $\partial_2$, the left-hand side becomes $(\sigma \times \mathrm{id})_\# ((\partial_2 H_1^{\Delta_1} + H_0^{\Delta_1}\partial_1)(\iota ))$. On the right-hand side, using $\sigma = \sigma_\# (\iota_1)$ and $i_j^X \sigma = (\sigma \times \mathrm{id}) i_j^{\Delta_1}$, we get $(i_j^X)_\# (\sigma ) = (\sigma \times \mathrm{id})_\# (\iota_1)$, so the right-hand side becomes $(\sigma \times \mathrm{id})_\# ((i_1^{\Delta_1})_\# (\iota_1) - (i_0^{\Delta_1})_\# (\iota_1))$.

Verify by an argument analogous to the one given above that

follows from the similar formula

Show that, if $f : X \to Y$ is a continuous map, then $H_i^Y(f_\# (\sigma )) = (f \times \mathrm{id})_\# (H_i^X(\sigma ))$, $i = 0, 1$ (with $H_i^X$ as in Exercises 6.16.1, 6.16.2, and 6.16.4).

We now extend $H_i^X(\sigma )$ to $i \geq 2$: first define it for $X = \Delta_i$ and $\sigma = [e_0, \ldots , e_i]$, showing it satisfies $\partial_{i+1} H_i([e_0, \ldots , e_i]) + H_{i-1}\partial ([e_0, \ldots , e_i]) = (i_1)_\# ([e_0,\ldots ,e_i]) - (i_0)_\# ([e_0,\ldots ,e_i])$, then extend via $H_i^X(\sigma ) = (\sigma \times \mathrm{id})_\# H_i^{\Delta_i}([e_0,\ldots ,e_i])$ (satisfying the analogous formula for $X$ and the same naturality as Exercise 6.16.5). We extend $H_i$ inductively, having already defined $H_j$ for $j < i$ on $\Delta_j$ and extended it over any $X$ via $H_j^X(\sigma ) = (\sigma \times \mathrm{id})_\# H_j^{\Delta_j}([e_0,\ldots ,e_j])$; in particular $H_{i-1}$ is already defined on the face maps $F_k : \Delta_{i-1} \to \Delta_i$, giving $H_{i-1}^{\Delta_i} \partial_i([e_0,\ldots ,e_i])$. Write $v_j = (e_j, 0)$ and $w_j = (e_j, 1)$ (points of $\Delta_i \times I$). The general definition of $H_i([e_0,\ldots ,e_i])$ in any dimension is

Show that

Show that

(notation as in Exercises 6.16.6-6.16.7).

Combine the calculations of Exercises 6.16.7 and 6.16.8 to show that

Show that if we define $H_i^X(\sigma ) = (\sigma \times \mathrm{id})_\# H_i([e_0,\ldots ,e_i])$ (with $H_i$ as in Exercises 6.16.6-6.16.9) and extend linearly to chains, then this satisfies the condition

Moreover, show that this satisfies the naturality property that when $f : X \to Y$, then $(f \times \mathrm{id})_\# H_i^X = H_i^Y f_\#$.

The formula of Exercise 6.16.10 shows that $(i_0^X)_\#$ and $(i_1^X)_\#$ are chain-homotopic chain maps, with chain homotopy $H_i^X$; hence they induce the same map in homology, $(i_0^X)_* = (i_1^X)_*$. Applying this to homotopic maps $f_0 \sim f_1$ via a homotopy $F : X \times I \to Y$: $(f_j)_* = F_*(i_j)_*$, so $(f_0)_* = (f_1)_*$, proving the homotopy property in the absolute case.

Prove that if $f : X \to Y$ is a homotopy equivalence, then $f_* : H_k(X) \to H_k(Y)$ is an isomorphism.