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...

>• 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...