# Epimorphisms in the category of partially ordered sets

This comment says that in the category of a partially ordered set, every arrow is an epi but no non-identity arrow has a right inverse.

My understanding is that the category in question is one where there is at most one arrow between any two objects, and there is an arrow $$a\to b$$ iff $$a\le b$$.

I'm having a trouble verifying the above claim (at least its second part).

For the first claim. Consider an arrow $$f:a\to b$$. To show it's an epi, we need to show that $$hf=h'f\implies h=h'$$ for all arrows $$h,h':b\to c$$. Well, suppose $$h,h':b\to c$$ are arrows such that $$hf=h'f$$. Since there is at most one arrow between $$b$$ and $$c$$, $$h=h'$$. This completes the proof. The assumption $$hf=h'f$$ is not even needed, right?

For the second claim. Suppose $$f:a\to b$$ is a non-identity arrow (this means $$a\le b$$). Assume it has a right inverse $$r:b\to a$$ (i.e., $$b\le a$$) for which $$fr=1_b$$ holds. What does it contradict to? I don't quite understand what statement $$fr=1_b$$ means in the language of $$\le$$.

• If $f\colon a\to b$ has a right inverse $r\colon b\to a$, then $a\leq b$ and $b\leq a$, hence $a=b$, and therefore $f$ is the identity, as that is the unique arrow from $a$ to itself. Thus “if an arrow has a right inverse, then it is the identity”, which is the contrapositive of “if it is not the identity, then it does not have a right inverse.” – Arturo Magidin Jul 26 '19 at 2:41
• @ArturoMagidin Is the proof of the first part in question correct? Do we indeed not need the assumption $hf=h'f$? – user634426 Jul 26 '19 at 2:50
• Yes; basically, because there can only be (at most) one map between any two objects, any implication that ends with asserting that two maps between a given pair of objects are equal is always true, and hence the implication is always true. – Arturo Magidin Jul 26 '19 at 3:11

In a partial order, if $$a \leq b$$ and $$b \leq a$$ then $$a = b$$ (this is antisymmetry). So it's impossible for there to be distinct elements with arrows in both directions (that would be a preorder). And of course, the unique arrow from an element to itself is the identity arrow.