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The singular cohomology of the loop space $\Omega S^3$ of the 3-sphere is a divided power algebra $\Gamma_{\mathbb Z}[s]$ on one generator $s$ of degree 2, so the rational cohomology is a the polynomial ring $\mathbb{Q}[s]$. Likewise, the singular cohomology of infinite complex projective space $\mathbb C \mathrm P^\infty = BS^1$ is a polynomial ring $\mathbb{Z}[t]$ on one generator $t$ of degree 2, so the rational cohomology is $\mathbb{Q}[t]$.

Since a symmetric algebra is a free CGA, any lifting of $s$ and $t$ to cocycles in CDGA models for the rational cohomology of these spaces induces a quasi-isomorphism between the cohomology ring, viewed as a CDGA with trivial differential, and the model, so these spaces are both formal. Moreover, both spaces are simply-connected. Thus any map between the two inducing an isomorphism in rational cohomology should be a rational homotopy equivalence.

Does there exist such a map?

P.S.: I guess the same question is equally (un)reasonable for $\Omega S^5$ and $\mathbb H \mathrm P^\infty$. What about this pair?

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There is a map $S^3\to K(\mathbb{Z},3)$ inducing an isomorphism on $\pi_3$. Looping this map, you get a map $\Omega S^3\to K(\mathbb{Z},2)=\mathbb{C}P^\infty$ inducing an isomorphism on $\pi_2$. It follows easily that this map is a rational equivalence.

For $\Omega S^5$ and $\mathbb{H}P^\infty$, I don't know whether you can get a map $\Omega S^5\to \mathbb{H}P^\infty$ which is a rational equivalence (the above approach doesn't work because you can't deloop $\mathbb{H}P^\infty$). However, there is no rational equivalence $\mathbb{H}P^\infty\to\Omega S^5$. For if you had such a map, then it would be nonzero on $H^4(-,\mathbb{Z})$, and so looking at the induced map on the whole integral cohomology rings, you would get a ring-homomorphism $\Gamma_\mathbb{Z}(s)\to\mathbb{Z}[t]$ sending $s$ to $nt$ for some nonzero integer $n$. But if $p$ is a prime not dividing $n$, such a homomorphism would have nowhere to send $s^p/p!$. (Alternatively, you could use Steenrod powers: a rational equivalence $\mathbb{H}P^\infty\to\Omega S^5$ would yield a map $\Sigma\mathbb{H}P^\infty\to S^5$ whose induced map on $H^5(-,\mathbb{Z})$ is multiplication by some nonzero integer $n$; reducing mod a prime not dividing $n$ and considering the Steenrod power $P^2$ then yields a contradiction). Note, however, that they can easily be connected by a zigzag of rational equivalences: for instance, a map from either of them to $K(\mathbb{Z},4)$ classifying a generator of $H^4$ is a rational equivalence.

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  • $\begingroup$ 1. Is there anything more geometric to be said about the map $\Omega S^3 \to \mathbb{C}\mathrm{P}^\infty$ in question? $\endgroup$ – jdc Dec 7 '15 at 2:51
  • $\begingroup$ 2. Where does the sufficient largeness of the prime come in? $\endgroup$ – jdc Dec 7 '15 at 3:04
  • $\begingroup$ 1. Here is one neat interpretation of it, which you may or may not consider "geometric". By a theorem of James, the space $\Omega S^3=\Omega\Sigma \mathbb{C}P^1$ can be identified (up to homotopy) with the free topological monoid on $\mathbb{C}P^1$ (as a pointed space, with the basepoint becoming the unit of the monoid), and $\mathbb{C}P^\infty$ can be identified with the free topological commutative monoid on $\mathbb{C}P^1$. This map $\Omega\Sigma \mathbb{C}P^1\to\mathbb{C}P^\infty$ is then just the canonical "abelianization" map. $\endgroup$ – Eric Wofsey Dec 7 '15 at 3:06
  • $\begingroup$ 2. You just need to choose a prime $p$ such that the induced map on $H^5(-,\mathbb{Z}/p)$ is nonzero. $\endgroup$ – Eric Wofsey Dec 7 '15 at 3:07
  • $\begingroup$ 3. I'm obviously missing something kind of major here, so I'd better ask. The cohomology of $\mathbb{H}P^\infty = BS^3$ should be a polynomial ring on one variable of degree four, and that of $S^4$ should be a truncated polynomial ring inhabited only in degrees zero and four. Both these spaces are simply-connected. How could it be that a map between these spaces could induce an isomorphism in rational cohomology? $\endgroup$ – jdc Dec 7 '15 at 3:09

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