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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 $\le$.

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    $\begingroup$ 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.” $\endgroup$ – Arturo Magidin Jul 26 '19 at 2:41
  • $\begingroup$ @ArturoMagidin Is the proof of the first part in question correct? Do we indeed not need the assumption $hf=h'f$? $\endgroup$ – user634426 Jul 26 '19 at 2:50
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    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Jul 26 '19 at 3:11
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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.

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