Let $X$ be a topological space. A map $f:X\longrightarrow X$ is called homotopy idempotent if $f\circ f\simeq f$. The set of all homotopy classes of homotopy idempotents over $X$ is denoted by $hI(X). $

Is $Card(hI(X\vee Y))$ determined by $Card(hI(X))$ and $Card(hI(Y))$?

  • 2
    $\begingroup$ Presumably you mean the set of homotopy classes of homotopy idempotents. $\endgroup$ Nov 27 '17 at 6:27
  • $\begingroup$ @QiaochuYuan Yes exactly. $\endgroup$
    – M.Ramana
    Nov 27 '17 at 6:34

This is not an answer. Let me restrict attention to Eilenberg-MacLane spaces $BG$ where $G$ is a discrete group. Then the monoid of homotopy classes of maps $BG \to BG$ can be identified with the set of conjugacy classes of endomorphisms $G \to G$, and the set of homotopy classes of homotopy idempotents can be identified with the set of conjugacy classes of "conjugation-idempotent" endomorphisms, meaning endomorphisms $F : G \to G$ such that there exists $h \in G$ such that

$$F^2(g) = h F(g) h^{-1} \forall g \in G.$$

Every such endomorphism is conjugate to an idempotent endomorphism, as follows: define $E(g) = h^{-1} F(g) h$. Then

$$E^2(g) = h^{-1} F(h^{-1} F(g) h) h = h^{-2} F^2(g) h^2 = h^{-1} F(g) h = E(g).$$

So in fact we only need to restrict our attention to conjugacy classes of idempotents. (This is a group-theoretic special case of the general fact that a map between two pointed path-connected spaces is homotopic to a based map.)

A similar calculation shows that two conjugate endomorphisms are in fact equal, so we only need to discuss idempotent endomorphisms. That is:

The set of homotopy classes of homotopy idempotent endomorphisms of an Eilenberg-MacLane space $BG$ can naturally be identified with the set of idempotent endomorphisms of $G$.

Idempotent endomorphisms of a group can be understood as follows. If $E : G \to G$ is any endomorphism, we always have a short exact sequence

$$\text{ker}(E) \to G \to \text{im}(E).$$

If $E$ is idempotent, the additional feature is that $\text{im}(E) = \text{fix}(E)$ is the group of fixed points of $E$, and in particular is a subgroup; said another way, $E$ equips the short exact sequence above with a canonical splitting $\text{im}(E) \to G$, exhibiting $G$ as the semidirect product of $\text{ker}(E)$ and $\text{im}(E)$. This is a natural bijection, hence:

The set of idempotent endomorphisms of a group $G$ can be identified with the set of pairs $(N, K)$ of subgroups of $G$ such that $G$ is the semidirect product $N \rtimes K$.

Call such a pair a splitting of $G$. Now, your question, specialized to EM spaces, asks the following purely group-theoretic question.

Is the number of splittings of the free product $G \ast H$ determined by the numbers of splittings of $G$ and of $H$?

I thought the answer would end up being clearly no and that it would be easy to write down a counterexample, but actually it's starting to seem like $G \ast H$ basically always has infinitely many idempotent endomorphisms, as long as $G$ and $H$ are both nontrivial, because the quotient map from $G \ast H$ to either $G$ or $H$ has lots of sections. The examples I looked at were all finitely generated so ended up all having countably many idempotent endomorphisms.

  • $\begingroup$ Thank you so much for your exact and complete explanation. This can be so useful for me because of enough information from group theoretical point of view. Thank you. $\endgroup$
    – M.Ramana
    Nov 27 '17 at 11:14
  • $\begingroup$ I was wondering if you could give me a counterexample about your last statement in your answer. Thank you. $\endgroup$
    – M.Ramana
    Jun 17 '18 at 11:30

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