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Let $X$ be a topological space. A continuous map $h:X\longrightarrow X$ is called homotopy idempotent if $h\circ h\simeq h$.

My question is:

What is the number of homotopy classes of homotopy idempotent maps $h:\prod_{n\in I}(\prod_{i_n}\mathbb{S}^n ) \longrightarrow \prod_{n\in I}(\prod_{i_n}\mathbb{S}^n )$, where $i_n$ denotes the number of copies of $\mathbb{S}^n$ (a finite product of spheres of the same or different dimensions), where $I$ is a finite subset of $\mathbb{N}$?

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  • $\begingroup$ Minor terminology question: should we consider multisets $I$, so that there can be more than one sphere in each dimension? $\endgroup$ – Joshua Mundinger Jun 16 '18 at 9:57
  • $\begingroup$ @JoshuaMundinger Your are right. I think it is better to write $\prod_{n\in I}(\prod_{i_n} \mathbb{S}^n)$, where $i_n$ denotes the number of copies of $\mathbb{S}^n$. $\endgroup$ – M.Ramana Jun 16 '18 at 12:32
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This seems difficult in general, since the homotopy classes of maps between spheres is a famous and unsolved problem. I like it!

A special case which is tractable (at least for me) is a self-map of the $n$-torus $f: (S^1)^n \to (S^1)^n$. Because the $n$-torus has contractible universal cover which is a group, any two endomorphisms of the $n$-torus which agree on $\pi_1$ are homotopic (their difference lifts to $\mathbb{R}^n$ and hence is nullhomotopic). Hence the homotopy endomorphisms of the $n$-torus (which have a group structure from the torus, making them into a ring with composition) are isomorphic as a ring to the ring of $n\times n$ integer matrices. There are infinitely many idempotent $n\times n$ integer matrices for $n>1$.

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