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Let $[X,S^1]$ denote the set of pointed homotopy classes of maps $f:X\to S^1$. I need to show that, when $S^1$ is viewed as a subset of $\mathbb{C}$, complex multiplication induces a group structure on $[X,S^1]$ such that the bijection $T:[X,S^1] \to H^1(X;\mathbb{Z})$ is an isomorphism. This bijection is given as Theorem 4.57 in Hatcher's Algebraic Topology; namely, given $f:X\to S^1$, the bijection is given by $T([f]) = f^*(\alpha)$, where $\alpha$ is a generator of $H^1(S^1;\mathbb{Z})$.

I know that $[S^1,S^1]$ is the same as $\pi_1(S^1)$, which already has a group structure; moreover, if $X$ is $S^1$ the operation of complex multiplication is the same as that on $\pi_1(S^1)$. The result holds for $X = S^1$, but I can't figure out how to generalize it.

Since $S^1$ is a $K(\mathbb{Z},1)$, I considered using the adjoint relation to show that $[X,S^1] = [\sum X, K(\mathbb{Z},2)]$, but I can't figure out where to go from here, either. Any help would be most appreciated.

Andrew
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    $S^1$ is already a subgroup of $\mathbb{C}$ and the group operation (multiplication) is continuous. So you can define a group strucuture on the set of continuous functions $f:X\to S^1$ for any X by multiplying functions pointwise. Then it is easy to see that this operation is also well defined on homotopy classes. – Arun Kumar Mar 18 '16 at 10:22

1 Answers1

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In general, $[X, K(G, n)]$ is a group whenever $G$ is abelian. This comes from the homotopy commutative, homotopy associative $H$-space structure on $K(G, n)$ which has inverses modulo homotopy.

This structure can be easily seen to come from the map $$K(G \times G, n) \simeq K(G, n) \times K(G, n) \to K(G, n)$$ induced from the group addition map $G \times G \to G$. This is homotopy associative and homotopy commutative since the group addition on $G$ is associative and commutative. The inverse map is the $K(-, n)$-level map induced from the map $G\to G$ defined by $x \mapsto x^{-1}$.

The group addition on $[X, K(G, n)]$ is defined by taking representatives $f, g : X \to K(G, n)$ of two chosen homotopy class $[f], [g]$, and defining the multiplication map $$f * g : X \to K(G, n) \times K(G, n) \to K(G, n)$$ where the first factor $X \to K(G, n) \times K(G, n)$ in the composition is defined by $x \mapsto (f(x), g(x))$. The addition is then $[X, K(G, n)] \times [X, K(G, n)] \to [X, K(G, n)]$ given by $([f], [g]) \mapsto [f*g]$. You can easily verify that this in fact satisfies the axioms of group addition.

In particular, restricting to the case $G = \Bbb Z$ and $n = 1$, you have a group structure on $[X, S^1]$.

(As a side-note, $[X, K(G, n)]$ is also in bijective correspondence with $H^n(X;G)$, and in fact you get an isomorphism of groups when $[X, K(G, n)]$ is given the structure of a group as defined above. The idea of the proof is to show that $[X, K(G, n)]$ is a reduced cohomology theory satisfying the dimension axiom. Then by classification of reduced cohomology theory, the isomorphism of functors $[-, K(G, n)]\cong H^n(-;G)$ follows)

Balarka Sen
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    Note that for the present case of $S^1 = K(\mathbb{Z}, 1)$, the circle is a group "on the nose" so of course $[X,S^1]$ is a group too. – Najib Idrissi Sep 04 '15 at 11:41