Homotopy invariance for cohomology

Question

I have tried to read Hatcher's Algebraic Topology book. He proved

Theorem 2.10: If two maps $f,g :X \to Y$ are homotopic, then they induce the same homomorphism $f_* = g_*:H_n(X) \to H_n(Y)$.

One can immediately obtain following corollaries using the dual version of prism operator described in Theorem 2.10 and some facts:

Corollary 1: If $f,g :X \to Y$ are homotopic, then $f_* = g_*:\tilde H_n(X) \to \tilde H_n(Y)$ for $n>0$.

Corollary 2: If $f,g :(X,A) \to (Y,B)$ are homotopic, then $f_* = g_*:H_n(X,A) \to H_n(Y,B)$.

Corollary 3: If $f,g :X \to Y$ are homotopic, then $f^* = g^*: H^n(X) \to H^n(Y)$.

Corollary 4: If $f,g :X \to Y$ are homotopic, then $f^* = g^*:\tilde H^n(X) \to \tilde H^n(Y)$ for $n>0$.

Corollary 5: If $f,g :(X,A) \to (Y,B)$ are homotopic, then $f^* = g^*:H^n(X,A) \to H^n(Y,B)$.

I wonder that how can we prove the following statements:

Corollary 6: If $f,g :X \to Y$ are homotopic, then $f_* = g_*:\tilde H_0(X) \to \tilde H_0(Y)$.

Corollary 7: If $f,g :X \to Y$ are homotopic, then $f_* = g_*:\tilde H^0(X) \to \tilde H^0(Y)$.

An attempt for Corollary 7 is

Proof of Corollary 7: By Universal Coefficient Theorem, we have $$\tilde H^0(X;G) \approx {\rm Hom}(\tilde H_0(X),G).$$

Hence the result follows from the dualization of the proof Corollary 6.

Any help will be appreciated.

Answer 1

As Hatcher says after Proposition 2.12, there are also induced homorphisms $f_* : \tilde{H}_n(X) \to \tilde{H}_n(Y)$ and the "same" proof as that of 2.10 shows that homotopic maps induce the same homomorphism. Recall that $\tilde{H}_n(X)$ is the $n$-th homology group of the augmented chain complex $$ \dots C_1(X) \stackrel{\partial_1}{\rightarrow} C_0(X) \stackrel{\epsilon}{\rightarrow}\mathbb{Z} \to 0 .$$ This is perhaps a little bit sloppy because after Lemma 2.1 Hatcher defines a chain complex to be a sequence of abelian groups ending with $0$ at position $-1$. Thus, formally, the augmented chain complex is not a chain complex because it does not end at position $-1$, nor has a $0$ at this position, but ends with $0$ at position $-2$. But I think it clear that we can define the $n$-th homology groups precisely as for an ordinary chain complex. Doing so, we see that for $n > 0$ we have $\tilde{H}_n(X) = H_n(X)$ (not only $\tilde{H}_n(X) \approx H_n(X)$).

For a map $f : X \to Y$ we get a homomorphism of augmented chain complexes by defining $f_\# = id : \mathbb{Z} \to \mathbb{Z}$ in position $-1$.

Now let $f, g : X \to Y$ be homotopic. The above comments show that $f_* = g_* : \tilde{H}_n(X) \to \tilde{H}_n(Y)$ for $n > 0$.

To adapt the proof of 2.10 for $n = 0$ we additionally need a prism operator $P_{-1} : \mathbb{Z} = C_{-1}(X) \to C_0(Y)$ such that $\partial P_0 = g_\# - f_\# - P_{-1} \epsilon$ as homomorphisms $C_0(X) \to C_0(Y)$. In fact $P_{-1} = 0$ will do. For a $0$-simplex $\sigma : \Delta^0 \to X$ we obtain the $1$-simplex $P_0(\sigma) = F \circ (\sigma \times \mathbb{I}) \mid [v_0,w_0]$. But then $$\partial P_0 (\sigma) = P_0(\sigma) \mid [v_0] - P_0(\sigma) \mid [w_0] = F \circ (\sigma \times \mathbb{I}) \mid [v_0] - F \circ (\sigma \times \mathbb{I}) \mid [w_0] \\ = f\circ \sigma - g \circ \sigma = (g_\# - f_\#)(\sigma) .$$

The cohomology part can be treated analogously.

Answer 2

There is no reason to reason so formally about this; it gives the impression that reduced homology requires as much careful thought as the other homology groups, which is not true. I will prove the statement for homology directly, and there is no difficulty in taking that direct proof to the case of cohomology.

The group $C_0(X)$ is the free abelian group on points $x \in X$. The group $C_1(X)$ is the free abelian group on paths $\gamma: [0,1] \to X$. The boundary $\partial [\gamma] \in C_0(X)$ is $[\gamma(1)] - [\gamma(0)]$. Setting this equal to zero amounts to setting $[\gamma(0)] = [\gamma(1)]$. If $x$ and $y$ are in the same path component, then there is some path connecting them, and hence $[x] = [y]$ in $H_0(X)$. There are no other relations than finite sums of these. When $X$ is path-connected, we thus have a natural isomorphism $H_0(X) \cong \Bbb Z$, sending $\sum c_i [x_i] \mapsto \sum c_i$; this is natural for continuous maps $f: X \to Y$ between path-connected spaces, as the sum-of-weights $\sum c_i [f(x_i)] \mapsto \sum c_i$ is exactly the same after pushing the points forward.

There is a map $H_0(X) \to \Bbb Z$ for each path-component given by sending $[x]$ to $1$ if it is in that path component, and otherwise to zero; this is a priori a map on $C_0(X)$, but $[x] - [y] \in \partial C_1(X)$ maps to $0$, because $x$ and $y$ are in the same path component. These assemble into a map $\rho: H_0(X) \to \Bbb Z^{\pi_0 X}$, where here the notation means $\Bbb Z^{\pi_0 X} = \bigoplus_{C \in \pi_0 X} \Bbb Z$. What we see from the previous paragraph is that this is an isomorphism by decomposing this into a direct sum of one map for each path-component.

Thus, $\rho$ gives an isomorphism $H_0(X) \cong \Bbb Z^{\pi_0 X}$, the free abelian group on the number of path components. This isomorphism is natural: given a map $f: X \to Y$, there is an induced map $\pi_0 X \to \pi_0 Y$, and hence an induced map $\Bbb Z^{\pi_0 X} \to \Bbb Z^{\pi_0 Y}$ which fits into a commutative diagram with the isomorpshisms $\rho_X$ and $\rho_Y$. (This is essentially the same computation as in the second paragraph, but now we say "for each path-component...")

Reduced homology is the kernel of the map $\pi: H_0(X) \to \Bbb Z$ which takes $\sum c_i [x_i] \mapsto \sum c_i$. It is thus naturally isomorphic to $\text{ker}(\pi \rho_X^{-1})$, where $\pi \rho_X^{-1}: \Bbb Z^{\pi_0 X} \to \Bbb Z$. The identifications of the above paragraph show that this composite is precisely the map $\Bbb Z^{\pi_0 X} \to \Bbb Z$ given by sending $\sum_{[C] \in \pi_0 X} c_i [C] \mapsto \sum c_i$ (here, recall that in the definition of a possibly-infinite direct sum of abelian groups, only finitely many of those terms $c_i$ are nonzero).

Hence, reduced homology is naturally isomorphic to the subspace of $\Bbb Z^{\pi_0 X}$ in which the sum of the coordinates is zero. In particular, the map $f_*: \tilde H_0(X) \to \tilde H_0(Y)$ is the restriction of the map $f_*: \Bbb Z^{\pi_0 X} \to \Bbb Z^{\pi_0 Y}$ to this subspace. This last map only depends on the induced map at the level of sets $f_*: \pi_0 X \to \pi_0 Y$, which is certainly a homotopy invariant of $f$, as desired.

The cohomological answer gives that $H^0(X;\Bbb Z)$ is the group of maps $\pi_1(X) \to \Bbb Z$ (for which infinitely many terms may be nonzero); equivalently it is $\prod_{\pi_1 X} \Bbb Z$. The reduced cohomology $\tilde H^0(X;\Bbb Z)$ gives the quotient space given by killing the constant function $f: \pi_1 X \to \Bbb Z$ with $f(C) = 1$ for all $C \in \pi_1 X$.