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I'm looking for a 'simple' path-connected space $X$ such that for some homotopy group $\pi_n(X)$ there are fewer free homotopy classes $[S^n,X]$ than elements of $\pi_n(X)$.

From this question we know that no such example can exist for $n=1$. EDIT: this is false! Thanks to Moishe Cohen. Ideally I would like an example for $n=2$.

Thanks for the help.

gj255
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    No, it does not follow and actually, such examples do exist for $n=1$. For instance, start with an infinite group $G$ which has exactly two conjugacy classes, see here (such examples exist even among finitely generated groups). Then let $X=K(G,1)$. Of course, maybe this $X$ does not qualify as "simple", but then you should explain what "simple" means. – Moishe Kohan Feb 14 '18 at 14:52
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    Whenever $G=\pi_1(X)$ is a finite nonabelian group then $G$ has fewer conjugacy classes than elements and so $X$ has fewer elements in $[S^1,X]$ than elements of $\pi_1(X)$. – Lee Mosher Feb 14 '18 at 14:57
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    You should clarify what "fewer" means, my interpretation is that you are comparing cardinalities. Then as an example for $n\ge 2$ take the wedge of $S^n$ with uncountably many $S^1$'s. – Moishe Kohan Feb 14 '18 at 16:45
  • @MoisheCohen Thanks for the comment. By 'fewer' I am indeed referring to cardinalities, although if I am allowed to be picky I would prefer $[S^n,X]$ to be finite. – gj255 Feb 14 '18 at 17:06

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Just to close this question:

Consider $n\ge 2$. Start with an $(n-1)$-connected space $Y$ such that $\pi_n(Y)$ is finite, $\cong {\mathbb Z}_k$, say, attach $n+1$-cell to $S^n$ along a degree $k$ map, $k>1$. Take $X= S^1\vee Y$.

Then $[S^n, X]\cong \pi_n(Y)\cong {\mathbb Z}_k$ is finite, while $\pi_n(X)$ is infinite, isomorphic to the countably infinite direct sum $$ \oplus_{i\in {\mathbb Z}} {\mathbb Z}_k.$$ The fundamental group of $X$ acts on this group as the shift.

Moishe Kohan
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As pointed out by the OP, this is equivalent to having a nontrivial action of $\pi_1(X)$ on $\pi_n(X)$.

Mikhail Katz
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  • I see, thank you. This seems to be in tension with this question (https://math.stackexchange.com/questions/1254666/bijection-between-the-free-homotopy-classes-sn-x-and-the-orbit-space-pi) however. – gj255 Feb 14 '18 at 16:34
  • Right. So then you answered your own question: take a space $X$ with a nontrivial action of the fundamental group on a higher homotopy group. – Mikhail Katz Feb 14 '18 at 16:42