$\langle X,S^1\rangle\to H^1(X;\Bbb Z)$ is an isomorphism

Question

(Hatcher exercise 4.3.2) Show that the group structure on $S^1$ coming from multiplication in $\Bbb C$ induces a group structure on $\langle X,S^1\rangle$ such that the bijection $\langle X,S^1\rangle\to H^1(X;\Bbb Z)$ of theorem 4.57 is an isomorphism.

Theorem 4.57 : there is a bijection $T:\langle X,S^1\rangle\to H^1(X;\Bbb Z)$ by $[f]\mapsto f^*(\alpha)$ where $\alpha\in H^1(S^1;\Bbb Z)$ is a generator.

I can define the group structure on $\langle X,S^1\rangle$ by defining $(f+g)(x):=f(x)\cdot g(x)$ where $f,g\in\langle X,S^1\rangle$. To show $T$ is an isomorphism, I need to show $f^*(\alpha)+g^*(\alpha) = (f+g)^*(\alpha)$. $f^*(\alpha)+g^*(\alpha) = [\alpha\circ f]+[\alpha\circ g] = [\alpha\circ f+\alpha\circ g]$ and $(f+g)^*(\alpha) = [\alpha\circ (f+g)] = [\alpha\circ (f\cdot g)]$. I can't see why $[\alpha\circ f+\alpha\circ g] = [\alpha\circ (f\cdot g)]$. Could you help?

I found some similar post but it's really talking about the group structure not isomorphism.

Answer 1

Let $f\times g\colon X\to T^2$ denote the map sending $x\mapsto (f(x),g(x))$. Let $m\colon T^2 \to S^1$ map $(a,b)\mapsto ab$ where $a,b$ are unit complex numbers.

Then as you have defined it, $f+g=m\circ (f\times g)$.

Thus $(f+g)^*(\alpha)=(f\times g)^* m^*(\alpha)$.

Let $p_1,p_2$ be projections $T^2\to S^1$ onto the first and second factors. Then $p_1(f\times g)=f$ and $p_2(f\times g)=g$.

Lemma We have $m^*(\alpha)=p_1^*(\alpha)+p_2^*(\alpha)\in H^1(T^2)$.

Proof: Note that $m$ maps a loop in $T^2$ going once round the first factor of $S^1$ to a loop going once round $S^1$. Similarly it maps a loop in $T^2$ going once round the second factor of $S^1$ to a loop going once round $S^1$.

Thus if a loop in $T^2$ goes $a$ times round the first factor of $S^1$ and $b$ times round the second factor of $S^1$, then $m^*(\alpha)$ will take the loop to one going $a+b$ times round $S^1$, and measure it against $\alpha$.

By contrast $=p_1^*(\alpha)$ would take the loop to one going $a$ times round $S^1$ and measure against $\alpha$, and $p_2^*(\alpha)$ would take the loop to one going $b$ times round $S^1$ and measure against $\alpha$.$\qquad \Box$

We have: \begin{eqnarray*}(f+g)^*(\alpha)&=&(f\times g)^* (p_1^*(\alpha)+p_2^*(\alpha))\\ &=&(f\times g)^*p_1^*(\alpha)+(f\times g)^*p_2^*(\alpha)\\ &=&(p_1(f\times g))^*(\alpha)+(p_2(f\times g))^*(\alpha)\\ &=& f^*(\alpha)+g^*(\alpha) \end{eqnarray*} as required.

Answer 2

Consider $f \cdot g: X \xrightarrow{f,g} S^1 \vee S^1 \xrightarrow{\cdot} S^1$. $(f \cdot g)^*$ is a map $H^1(S^1) \to H^1(S^1) \times H^1(S^1) \to H^1(X).$ The second map in this composition is $(1,0) \mapsto f^*(\alpha), (0,1) \mapsto g^*(\alpha)$ (I see $H^1(S^1) \times H^1(S^1)$ as $\mathbb{Z} \times \mathbb{Z}$ but remember that $H^1(S^1)$ is generated by $\alpha$). So you are only left to show that the first map works as $\alpha \mapsto (1,1).$ You can use cellular chains, for example. The corresponding map $C_1(S^1) \times C_1(S^1) \to C_1(S^1)$ looks like $(a,b) \mapsto a+b.$ What is dual to the generator of $C_1(S^1)$ from the right will be dual to generators of multiples on the left when pulled back via $-\cdot -.$