Let $X$ denote an infinite set. Is every partitioning of $X^2$ induced by an associative operation $f:X^2 \rightarrow X$?

Proposition. Let $X$ denote an infinite set. Then for each partitioning $\Pi$ of $X$, there exists a function $f : X \rightarrow X$ whose coimage is $\Pi$.

I'd like to know whether the analogous statement holds for associative binary operations.

Question. Let $X$ denote an infinite set. Is it true that for each partitioning $\Pi$ of $X^2$, there exists an associative operation $f : X^2 \rightarrow X$ whose coimage is $\Pi$?

This question occurred to me as a result of Behrouz Maleki's answer here.

• You can't expect all partitions of a large set to be given as the coimage of a function to a smaller set. In particular, take $X$ to have two elements, and the discrete partition on $X\times X$. This partition is not the coimage of any function $X^2 \to X$. Am I missing something? – Ittay Weiss Jun 5 '16 at 7:27
• @IttayWeiss, you're completely right, but I'm assuming that $X$ is infinite. I've edited to emphasize this. I like the idea of focusing on the discrete partition, though. – goblin Jun 5 '16 at 7:29

Consider the discrete partition of $X^2$, and assume it is the coimage of an associative binary operation. Then for all $a,b,c\in X$ we have $a(bc)=(ab)c$, and thus $a=ab$, and this determines the operation as projection to the left argument. Similarly, $bc=c$, and this determines the operation as projection to the right argument. This can only happen if $X$ is empty or a singleton.
• Good thinking! $\;\!$ – goblin Jun 5 '16 at 8:36