Why is $[\bigvee_IS^n,\bigvee_JS^n]_*\cong \operatorname{Hom}_{\mathbb{Z}}({\mathbb{Z}}[I],{\mathbb{Z}}[J])$

Question

How to show $[\bigvee_IS^n,\bigvee_JS^n]_*\cong \operatorname{Hom}_{\mathbb{Z}}({\mathbb{Z}}[I],{\mathbb{Z}}[J])$, where ${\mathbb{Z}}[S]$ denotes the free abelian group generated by a set $S$?

Answer 1

First, observe that the wedge product is the coproduct in the category of based topological spaces. That is, $Map_*(X\vee Y,Z)\cong Map_*(X,Z)\times Map_*(Y,Z)$. This holds for arbitrary wedges of spaces, and moreover also holds in the homotopy category, so that

$[\bigvee_IS^n,\bigvee_JS^n]\cong\prod_I[S^n,\bigvee_JS^n]$

with the bijection being given by sending $f:\bigvee_IS^n\rightarrow \bigvee_JS^n$ to the element $(f\circ in_i)_{i\in I}$, where $in_i:S^n\hookrightarrow \bigvee_IS^n$ is the inclusion of the ith wedge summand.

Observe that it makes no difference if we work in the based or unbased settings here.

Now for $n>1$ study the homology exact sequence of the pair $(\prod_JS^n,\bigvee_JS^n)$, and conclude that the inclusion $\bigvee_JS^n\hookrightarrow \prod_JS^n$ is a $(2n-1)$-equivalence. Therefore the inclusion induces bijections

$[S^n_i\bigvee_JS_j^n]\cong[S^n_i,\prod_JS_j^n]\cong \prod_J[S^n_i,S_j^n]$

which sends $f_i:S^n_i\rightarrow \bigvee_JS_j^n$ to the element $(q_j\circ f_i)_{j\in J}$ where $q_j:\bigvee_JS^n_j\hookrightarrow \prod_jS^n_j\xrightarrow{pr_j} S^n_j$ is the pinch map. (We'll introduce labels $i\in I$ and $j\in J$ at this point to help avoid confusion).

It follows that

$[\bigvee_IS^n_i,\bigvee_JS^n_j]\cong\prod_I[S^n_i,\bigvee_JS^n_j]\cong\prod_I \prod_J[S^n_i,S_j^n]$

given by $f\mapsto (f_{ij}=q_j\circ f\circ in_i)_{i\in I,j\in J}$.

Now observe that each $f_{ij}$ is a homotopy class in $[S^n_i,S^n_j]=\pi_n(S^n_j)\cong\mathbb{Z}\cong Hom_\mathbb{Z}(\mathbb{Z}(i),\mathbb{Z}(j))$, where the last bijection comes from sending $\alpha:S^n_i\rightarrow S^n_j$ to $\alpha_*:\pi_n(S^n_i)\rightarrow \pi_n(S^n_j)$ (we'll write $\mathbb{Z}(i)$ for the copy of $\mathbb{Z}$ with generator $i$).

Make the above identification and define a map $\Phi:[\bigvee_IS^n_i\bigvee_JS_j^n]\rightarrow Hom_\mathbb{Z}(\mathbb{Z}[I],\mathbb{Z}[J])$ by sending $f$ to the matrix $(f_{ij*})_{i\in I, j\in J}$. You can check that this is a bijection by producing the inverse, which we will do as follows.

A homomorphism $\varphi:\mathbb{Z}[I]\rightarrow\mathbb{Z}[J]$ is completely determined by what it does on each generator $i\in I$. In turn this is completely determined by the projection $\varphi_{ij}=pr_j(\varphi(1\cdot i))$ onto each of the $j\in J$ summands in the target.

Now choose a representing class $k:S^n\rightarrow S^n$ of the degree $k$ map, for each $k\in\mathbb{Z}$ and let $\Theta:Hom_\mathbb{Z}(\mathbb{Z}[I],\mathbb{Z}[J])\rightarrow [\bigvee_IS^n_i\bigvee_JS_j^n]$ be given by $\Theta(\varphi)=(\varphi_{ij})$ where $\varphi_{ij}$ is the unique homotopy class determined by $S^n_i\xrightarrow{in_i}\bigvee_IS^n_i\xrightarrow{\varphi(1\cdot i)} \bigvee_JS^n_j\xrightarrow{pr_j}S^n_j$ the composition.

I'll leave you to check that these are inverse bijections. Another thing you might like to check is the compatibility with composition. That is, given $\bigvee_KS^n_k$ we get a composition map

$\circ:[\bigvee_J S^n,\bigvee_KS^n]\times[\bigvee_I S^n,\bigvee_JS^n]\rightarrow [\bigvee_I S^n,\bigvee_KS^n]$

and you should check that this is just the matrix multiplication

$Hom_\mathbb{Z}(\mathbb{Z}[J],\mathbb{Z}[K])\times Hom_\mathbb{Z}(\mathbb{Z}[I],\mathbb{Z}[J])\rightarrow Hom_\mathbb{Z}(\mathbb{Z}[I],\mathbb{Z}[K]).$