Anindhya wrote :
>• if a finite non-empty poset is *discrete*, we can make it monoidal by slapping any old monoid structure on it (eg turn it into a cyclic group) – since \$$x\leq y \iff x = y\$$ this monoid automatically respects the partial order

>• if a poset has *all finite joins*, we can make it monoidal by choosing the bottom element (ie the join of nothing!) to be the unit and the binary join \$$x\vee y\$$ to be the monoidal product

>• if a poset has *all finite meets*, we can make it monoidal by choosing the top element (ie the meet of nothing!) to be the unit and the binary meet \$$x\wedge y\$$ to be the monoidal product

>Intuitively I think the best way of looking at this is that the partial order is too irregular to admit a monoidal structure: \$$a\$$ and \$$b\$$ don't have a join, \$$c\$$ and \$$d\$$ don't have a meet, there is no top element, nor is there a bottom one... basically \$$\leq\$$ is too badly behaved for any \$$\otimes\$$ operation to respect it.

Thank you! This made it clearer in my head. So pretty much the preorder needs to have a top or bottom element in order to define a well behaved identity for the monoid.

You also wrote:
>we're assuming here that \$$a, b, c, d\$$ are all mutually distinct (there's no "collapsing" going on).

>If you set \$$a = I\$$ then you can prove that \$$c\otimes d \leq c\$$ and \$$c\otimes d \leq d\$$.

What I meant by collapse was this observation which may or may not be valid.
Since \$$c\otimes d \leq c\$$ and \$$c\otimes d \leq d\$$, the only way this can be true is if c and d are both identities. Therefore, If you set \$$a = I\$$, then for the bowtie, \$$c = d = I\$$ and the set \$$a, b, c, d\$$ collapses down to \$$b, I\$$. So the contradiction is that a,c,d are not distinct.

Either way thanks for the great explanation.

Edit: Oops I said something completely different when I said collapse in comment 21. I think I got confused trying to understand both Matthew and Anindya's answers at once...