For any set $A\neq\varnothing$ let $\text{End}(A)$ denote the endomorphism monoid, consisting of all functions $f:A\to A$, together with composition. If $A, B\neq \varnothing$ are sets and $g:B\to A$ is a surjection, is there a surjective monoid homomorphism $\varphi:\text{End}(B)\to \text{End}(A)$?
$\begingroup$
$\endgroup$
6
asked Jul 10, 2020 at 7:02
-
$\begingroup$ We may take $B$ as a disjoint union of $A$ and another set $C$. Then for each endomorphism of $A$ we have an extension to $B$ by requiring the map is the identity when restricted to $C$. So the morphism $\varphi$ even splits. Edit: I took it for granted that the map $\varphi$ is already well-defined. I have to think more about it. I think I should keep this incorrect answer. $\endgroup$– GoulagCommented Jul 10, 2020 at 7:10
-
1$\begingroup$ Malcev proved that for $X$ infinite, $\mathrm{End}(X)$ has no nontrivial finite quotient, so the answer is no in case $B$ is infinite and $A$ is finite of cardinal $\ge 2$. Probably it's also no for $B$ finite and large enough and $1<|A|<|B|$, by the knowledge of normal congruences of $\mathrm{End}(B)$ (other users here know more about this and might confirm of infirm). $\endgroup$– YCorCommented Jul 10, 2020 at 7:54
-
2$\begingroup$ The question is actually unrelated to the original surjection, just the assumption is (in ZFC) that $|B|\ge |A|>0$. $\endgroup$– YCorCommented Jul 10, 2020 at 7:55
-
$\begingroup$ It seems that $|B|=3$ and $|A|=2$ is already an interesting case. Perhaps one could develop a program to test (when $|B|$ is finite). $\endgroup$– GoulagCommented Jul 10, 2020 at 8:11
-
1$\begingroup$ Probably not. Every monoid homomorphism $f:End(\omega_1)\to End(\omega)$ is trivial. Indeed $f$ induces $f:Sym(\omega_1)\to Sym(\omega)$. Every nontrivial quotient of $Sym(\omega_1)$ contains an uncountable direct sum of nonabelian groups (consequence of Baer), and this can't be mapped injectively in $Sym(\omega)$ (MacKenzie). Hence $f$ is trivial (= constant $\mathrm{id}_\omega$)) on $Sym(\omega)$. I'm not sure how to conclude; this probably follows from Malcev's classification of quotients of $End(X)$. $\endgroup$– YCorCommented Jul 10, 2020 at 9:37
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
1
This is a cleaner rewrite of my original answer. The answer is no (assuming the surjection is not injective and the smaller set does not have cardinality $1$).
Let $T_A$ be the full transformation monoid on the set $A$. Then the set $C_A$ of constant maps is the unique minimal two-sided ideal of $T_A$. Since $C_A$ has the same cardinality as $A$, we have $T_B\cong T_A$ if and only if $A$ and $B$ have the same cardinality.
I claim if $|A|\geq 2$, then the unique minimal non-trivial congruence on $T_A$ is to identify all the constant maps to a zero element (absorbing element). Assuming $T_A$ acts on the left of $A$, we have that $T_A$ acts faithfully on the left of $C_A$ by essentially the same action. So any homomorphism that is injective on $C_A$ is injective on $T_A$. On the other hand, if a congruence identifies elements of $C_A$ then the restriction of the congruence to $C_A$ is a system of imprimitivity for the symmetric group $S_A\leq T_A$ acting on the left of $C_A$ which is just the same as its natural action on $A$. This action is $2$-transitive and hence primitive. Thus any non-trivial congruence on $T_A$ must collapse $C_A$.
In conclusion, every proper quotient of $T_A$ has an absorbing element and so can only be a $T_X$ if $|X|=1$. Combined with the fact that $T_A\cong T_B$ iff $A$ and $B$ have the same cardinality, we get the answer is no.
answered Jul 10, 2020 at 17:09
-
1$\begingroup$ Very nice answer, thank you Benjamin! $\endgroup$ Commented Jul 12, 2020 at 8:50