Lifting properties of morphisms in category of sets

Question

Let $f:A\to B,g:C\to D$ be morphisms in a category $\mathcal C.$ We denote $f\perp g$ if in every commutative square $$\require{AMScd} \begin{CD} A @>>> C\\ @VfVV @VVgV \\ B @>>> D \end{CD}$$ there is a unique $d:B\to C$ making both triangles commute. Moreover, for a class $X\subseteq\operatorname{Mor}(\mathcal C)$ denote $X^\perp=\{g\mid\forall f\in X:f\perp g\},{}^\perp X=\{f\mid\forall g\in X:f\perp g\}$. Now,I want to show that in $\mathbf{Set}$ we have $Epi^\perp=Mono$. I think this should be somehow easy but I got lost in universal quantifiers. Inclusion "$\subseteq$" should follow from $$\require{AMScd} \begin{CD} A @>u>> C\\ @V{\text{id}}VV @VVgV \\ A @>>> D \end{CD}$$ where the unique diagonal is $u$ $\color{blue}{\text{(correct?)}}$ but what about the other? I don't even know what exactly I should prove. Given a mono $g:C\to D$ such that every commutative square $$\require{AMScd} \begin{CD} \bullet @>>> C\\ @VVV @VVgV \\ \circ @>>> D \end{CD}$$ has a unique diagonal $\circ\to C,$ then (every) $\bullet\to\circ$ is an epi?

Answer 1

I wasn't able to make your first diagram work. Instead here is an idea from @DanielSchepler. If you have two morphisms $f_1$, $f_2:A\to C$ such that $gf_1 = gf_2$, then consider the commuting diagram $$\require{AMScd} \begin{CD} A\sqcup A @>f_1\sqcup f_2>> C\\ @V\lambda VV @VVgV \\ A @>gf_1=gf_2>> D \end{CD}$$ where $\lambda$ is the map which "forgets" which side of the disjoint union a given $x\in A$ came from.

By assumption, this has a unique diagonal $d:A\to C$ for which the diagram still commutes. The top triangle commutes, i.e. $f_1\sqcup f_2 = d\circ\lambda$. By looking at the two halves of $A\sqcup A$, we can see that $f_1 = d = f_2$.

Conversely, if $g$ is a mono, consider an epi $f$ in the commuting diagram $$\require{AMScd} \begin{CD} A @>u_1>> C\\ @VfVV @VVgV \\ B @>u_2>> D \end{CD}$$ The claim is that regardless of the choice of $f$, $u_1$, $u_2$ here, there exists a unique $d:B\to C$ such that the diagram still commutes when $d$ is added. (This is different from your interpretation of the $\supseteq$ direction! If you want to try answering the problem yourself, starting from the correct statement, stop reading here.)

Finally, we construct $d$. Up til now we haven't actually used any properties of Set; now we need to. Consider that $\text{Im}(u_2) \subseteq \text{Im}(g)$ as subsets of $D$; this must be true since otherwise the diagram couldn't commute (and because $f$ is epi). But then since $g$ is mono we can take $g^{-1}:\text{Im}(u_2)\to C$ as a well-defined function, and write $d = g^{-1}u_2$. The fact that the resulting diagram commutes is pretty-much immediate from this definition of $d$.