# Surjective monoid homomorphism $\text{End}(B)\to \text{End}(A)$ given surjection $g:B\to A$

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 $$\varphi:\text{End}(B)\to \text{End}(A)$$?

• 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). – YCor Jul 10 '20 at 7:54
• The question is actually unrelated to the original surjection, just the assumption is (in ZFC) that $|B|\ge |A|>0$. – YCor Jul 10 '20 at 7:55
• Thanks @YCor for your comments! Is $\text{End}(\omega)$ a quotient of $\text{End}(\omega_1)$? – Dominic van der Zypen Jul 10 '20 at 8:23
• 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 $End(X)$. – YCor Jul 10 '20 at 9:37

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.

• Very nice answer, thank you Benjamin! – Dominic van der Zypen Jul 12 '20 at 8:50

As I cannot comment now, I write it as an answer. 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.

• 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). – Goulag Jul 10 '20 at 8:11

At least if there is an isomorphism $$g : A \rightarrow B$$, then there is an isomorphism given by $$F(h) = g \circ h \circ g^{-1} : B \rightarrow B$$, if $$h : A \rightarrow A$$ is isomorphism. Now if $$g$$ is only surjective, to construct the inverse this approach would need the axiom of choice. It might be possible with some other approach without choice, but I suspect more complex argument is needed. Whether this is a monoid homomorphism is left as execise for the reader 😃 More details on axiom of choice are in "Lawvere, Rosebrugh: Sets for Mathematics".