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Let $X$ be a set and let $(X^X,\circ)$ denote the monoid of all maps $f: X\to X$, together with composition. Let $(\text{Sym}(X),\circ)$ be the group of all bijections from $X$ to itself.

Does there exist a monoid homomorphism $h:X^X \to \text{Sym}(X)$ such that for every group $G$ and every monoid homomorphism $f: X^X\to G$ there is a homomorphism $f': \text{Sym}(X)\to G$ such that $f = f'\circ h$?

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  • $\begingroup$ No. For example the identity map $X^X \to X^X$ don't factor through $Sym(X)$ for a finite set $X$ because $Sym(X)$ has less elements than $X^X$. $\endgroup$
    – S. carmeli
    Commented Apr 23, 2018 at 6:40
  • $\begingroup$ $X^X$ is not a group in your example (I require $G$ above to be a group). $\endgroup$
    – Dominic van der Zypen
    Commented Apr 23, 2018 at 6:41
  • $\begingroup$ oh, I see, sorry. $\endgroup$
    – S. carmeli
    Commented Apr 23, 2018 at 6:46
  • $\begingroup$ You are asking the backward universal property. If M is a monoid, then its group of units is the universal group with a morphism INTO $M$. So any homomorphism from a group $G$ into $M$ factors through $Sym(X)$. If a monoid contains a right or left zero, all its group images are trivial. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 23, 2018 at 13:54
  • $\begingroup$ BTW, of you ask for uniqueness of $f'$ the answer is no because for the trivial homomorphism $X^X$ to $Sym(X)$ you have two extensions: the identity and the trivial homomorphism. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 23, 2018 at 13:57

1 Answer 1

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Yes for trivial reasons. Let $c$ be a constant map. Then for any two $f$ in $X^X$ we have $ c \circ f = c$. Hence the image of $f$ under any homomorphism to a group must be trivial.

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    $\begingroup$ By the way -- why have you both answered and voted to close? $\endgroup$
    – Stefan Kohl ♦
    Commented Apr 23, 2018 at 12:32
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    $\begingroup$ I agree with Stefan, especially insofar as the answer easily fits in a comment. $\endgroup$
    – YCor
    Commented Apr 23, 2018 at 12:52
  • $\begingroup$ @StefanKohl For a few reasons: (1) When I first saw the question, I voted to close. When I saw it had an upvote and no other close votes, I thought maybe it wouldn't be closed, and thus it would be better to be answered than to remain an unanswered question. (2) When I first started typing the answer, it was a little longer (I thought the argument would have to first show every function was equivalent to a constant function, then show all constants are equivalent), not all at once. (3) I knew Dominic was unlikely to rephrase the question after seeing it answered in the comments. $\endgroup$
    – Will Sawin
    Commented Apr 23, 2018 at 14:05
  • $\begingroup$ @YCor That said, I would be willing to delete this answer. $\endgroup$
    – Will Sawin
    Commented Apr 23, 2018 at 14:06

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