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



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


