Prove that every discrete poset is projective.

Question

Is this proof correct (assuming the Axiom of Choice)?

$e:P\longrightarrow Q$ epic, then $e$ is a surjection (Proof that the epis among posets are the surjections). For any arrow, $f:A \longrightarrow Q$, since $e$ is a surjection, there exists $p_{f(a)} \in P$ such that $e(p_{f(a)})=f(a)$ for any $a\in E$.

Assuming the Axiom of Choice, we can define: $ h: A \longrightarrow P : a \longmapsto p_{f(a)} $ Since the elements in $A$ are incomparable ($A$ is discrete), $h$ is a monotone map.

Hence we have $e\circ h=f$, $A$ is projective.

I also want to prove the inverse. Any idea how I could do that?

Thanks!

Answer 1

Your proof in the forward direction is correct. For the reverse direction, suppose $a<b$ for some $a,b\in A$. Then let $Q$ be the poset $\mathbf 2$ (where $0<1$), and define $f:A\to Q$ so that $f(x)=1$ if $a<x$ and $f(x)=0$ otherwise. If $x\le y$, then either $f(x)=0\le f(y)$, or $f(x)=1$ so $a<x\le y$, hence $f(y)=1\ge f(x)$. Thus $f$ is a monotone map.

Now take $P=\{0,1\}$ as a discrete poset, and $e$ be the identity map $P\to Q$. It is surjective, hence an epi by your linked theorem. Now let $h:A\to P$ satisfy $e\circ h=f$. Then $0=f(a)=e(h(a))=h(a)$, and $0=f(b)=e(h(b))=h(b)$, but $a\le b$ and $0\not\le 1$ in $P$, a contradiction.

Answer 2

Your proof is good. Here is another similar proof using your notaion, since $e$ is a surjective there is (according to the axiom of choice) a map $s:Q\to P$ such that $es=1_Q$. It follows that $e(sf)=f$ and since $A$ is discrete $sf$ is a monotone map proving that $A$ is projective.

For the converse suppose $A$ is projective and let $A'$ be the discrete poset on the same set. Let $e:A'\to A$ be the identity map (which is trivially a monotone epimorphism) and let $f: A\to A$ be the identity map. Since $A$ is projective there is a monotone map $h:A\to A'$ such that $eh=f$. It follows that $h$ is necesserily the identity map too and hence $A$ is discrete.