This is probably a very basic question, but I don't know the answer and I also don't see how it might be obvious (which it very well might be).


Given a topological space $X$, when is the set of all its continuous self-maps generated by the subset consisting of


* the self-homeomorphisms of prime order (i.e., there exists some prime number $p$ such that $f^p = id$ while $f^k \neq id$ for all $k \leq p$) and
* the continuous retractions (i.e. $f^2 = f$)?


Is the answer to this a well-known fact? Is there a characterization of the cases where it is true?


I should add that, by generation, I mean everything you can get by applying composition finitely many times.




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Edit: I originally also asked whether the statement is true or not. A counterexample was quickly presented by James Cranch (see below), so I have put more emphasis on a different part of my question, namely if there is a characterization of the cases where the statement is true.



When are all continuous self-maps of a topological spaces generated by retractions and self-homeomorphisms of prime order?
This is most likely pretty rare. Let G be any finite group. Then Birkhoff proved that G is the automorphism group of a finite distributive lattice L. The G is the group of homomorphisms of the space of prime filters on L with the usual kernel-hull topology. Thus any finite group is the homeomorphism group of a finite topological space but need not be generated by elements of prime order. 


**Added.**In fact any group is the homeomorphism group of a topological space, including torsion-free ones and any monoid can be the set of nonconstant selfmaps of a space. See [What sets of self-maps are the continuous self-maps under some topology?](https://mathoverflow.net/questions/136119/what-sets-of-self-maps-are-the-continuous-self-maps-under-some-topology). 


