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