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This question is pretty much this question stated in slightly different way. All projective spaces are complex ones. Let's assume $X$ is a connected CW complex. We have a natural map in the following form: $$f: Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty}))\rightarrow Sym^{\infty}(Hom(X, \mathbb{P}^{\infty}))\rightarrow Hom(X, Sym^{\infty}(\mathbb{P}^{\infty}))$$

Superscript zero denotes the connected component corresponding to zero. Since $\mathbb{P}^{\infty}\cong K(\mathbb{Z}, 2)$ so $\pi_0(Hom(X, \mathbb{P}^{\infty}))=H^2(X, \mathbb{Z})$.

By Dold-Thom $\pi_i(Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty})))=H_i(Hom^0(X, \mathbb{P}^{\infty})))$. There is a $H$-space structure on $Hom^0(X, \mathbb{P}^{\infty})$ induced from $\mathbb{P}^{\infty}$. This $H$-space structure turns $H_i(Hom^0(X, \mathbb{P}^{\infty})))$ into a graded algebra. Consequently $\pi_i(Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty})))$ is a graded algebra.

  • Does $f$ map this algebra product to the cup product on the right side?

Note that homotopy groups of the right side is as the following:

We have $Sym^{\infty}(\mathbb{P}^{\infty})=\prod_{i=1}^{\infty}K(2i, \mathbb{Z})$ so $\pi_i(Hom(X, Sym^{\infty}(\mathbb{P}^{\infty})))=\bigoplus H^{\text{even}}(X, \mathbb{Z})$ if $i$ is even and $\bigoplus H^{\text{odd}}(X, \mathbb{Z})$ if $i$ is odd. So there is a natural cup product structure on the right.

Let's assume as Tom Goodwillie mentions in the comments that the homotopy equivalence between $Sym^{\infty}(\mathbb{P}^{\infty})$ and $\prod_{i=1}^{\infty}K(2i, \mathbb{Z})$ is induced by mapping $\mathbb{P}^{\infty}$ to $\prod_{i=1}^{\infty}K(2i, \mathbb{Z})$ where each map $\mathbb{P}^{\infty}\rightarrow K(2i, \mathbb{Z})$ is the map in homotopic to the one corresponding to the generator of $H^{2i}(\mathbb{P}^{\infty}, \mathbb{Z})$. Now the map from $\mathbb{P}^{\infty}$ to $\prod_{i=1}^{\infty}K(2i, \mathbb{Z})$ is extended to a map from $Sym^{\infty}(\mathbb{P}^{\infty})$ to $\prod_{i=1}^{\infty}K(2i, \mathbb{Z})$ by the $H$-space operation. So the final map is a map of $H$-spaces.

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  • $\begingroup$ To have a definite question one must specify a particular homotopy equivalence to the product of $K(2i,\mathbb Z)$. $\endgroup$ Jul 8, 2022 at 23:27
  • $\begingroup$ @TomGoodwillie Does choosing different homotopy equivalences lead to different maps on homotopy groups? I don't know an explicit homotopy equivalence but it is in the Dold-Thom paper. $\endgroup$
    – user127776
    Jul 9, 2022 at 0:34
  • $\begingroup$ Yes, it definitely does. $\endgroup$ Jul 9, 2022 at 17:38
  • $\begingroup$ But doesn't it change the morphism up to the automorphisms of the cohomology groups and automorphisms respect the cup product. What you are saying is like saying different models for Eilenberg Maclane spaces induce different cup products. $\endgroup$
    – user127776
    Jul 9, 2022 at 17:55
  • $\begingroup$ Let's specify that the equivalence from $Sym^\infty(\mathbb P^\infty)$ to the product of the $K(2i,\mathbb Z)$ should be an $H$-space map, so that it is determined by its restriction to $\mathbb P^\infty$. And let's specify that the map $\mathbb P^\infty\to K(2i,\mathbb Z)$ should be the one that corresponds to the usual generator of $H^{2i}(\mathbb P^\infty)$. (There might be a sign ambiguity here, but I guess we know what we mean by the usual one.) In that case, I withdraw my objection. $\endgroup$ Jul 9, 2022 at 19:06

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No, not quite.

Let's set the stage like this:

If $G$ is a connected $H$-space then $Sym^\infty G$ is a ring space, so that $Hom(X,Sym^\infty G)$ is also a ring space and $\pi_\ast Hom(X,Sym^\infty G)$ is a graded ring.

Also, $Hom^0(X,G)$ is a connected $H$-space, so $Sym^\infty Hom^0(X,G)$ is a ring space and $\pi_\ast Sym^\infty Hom^0(X,G)$ is a graded ring.

The canonical map $Sym^\infty Hom^0(X,G)\to Hom(X,Sym^\infty G)$ is a ring space map. Therefore $\pi_\ast Sym^\infty Hom^0(X,G)\to \pi_\ast Hom(X,Sym^\infty G)$ is a map of graded rings.

Now what does this have to do with cup products?

I would say that the graded ring $\pi_\ast Hom(X,Sym^\infty G)$ is the cup product ring for cohomology of $X$ with coefficients in the graded ring $\pi_\ast (Sym^\infty G)=H_\ast (G)$.

That is, the multiplication $$ H^i(X;H_{2j}(\mathbb P^\infty))\times H^k(X;H_{2\ell}(\mathbb P^\infty))\to H^{i+k}(X;H_{2j+2\ell}(\mathbb P^\infty)) $$ is a cup product. But it's a cup product based on the multiplication $$ (*)\ \ \ H_{2j} \mathbb P^\infty\times H_{2\ell}\mathbb P^\infty\to H_{2j+2\ell}\mathbb P^\infty $$ (induced by the $H$-space structure). This cannot be identified with the cup product $$ H^i(X;\mathbb Z)\times H^k(X;\mathbb Z)\to H^{i+k}(X;\mathbb Z), $$ because the map $(\ast)$ above does not take generators to generators; there is a factor of ${j+\ell}\choose {j}$.

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  • $\begingroup$ Sorry for getting back to this after a long time, I came across something that always thought it is trivial but then realized it isn't. Why infinite symmetric product of a connected $H$-space is a ring space? Ring space I assume you mean there is a smash product (which turns homotopy groups into a ring). But I don't see that. I'm aware that pairing of the points on the symmetric product and summing them using the $H$-space structure gives a new pairing but it doesn't give a smash pairing. $\endgroup$
    – user127776
    Apr 22, 2023 at 23:08

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