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. $\endgroup$ – Qiaochu Yuan 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.