Condition for partial maps being representable

Question

Let $\operatorname{Part}(-,Y)$ be the presheaf which takes an element $X$ to the set $\operatorname{Part}(X,Y)$ of partial maps into $Y$. We say that partial maps are representable if $\operatorname{Part}(-, Y)$ is representable for each $Y$. I want to show that this is equivalent to the usual definition with the partial map classifier. Namely: for each $Y$ there exists a monomorphism $\eta_Y\colon Y\to\bar{Y}$ such that for any partial map $X\leftarrow U\xrightarrow{f} Y$ there is a unique map $\bar f\colon X\to\bar{Y}$ such that $$ \require{AMScd} \begin{CD} U @>{f}>> Y\\ @VVV @VVV \eta_Y \\ X @>{\bar{f}}>>\bar{Y} \end{CD} $$ is a pullback square.

Answer 1

The Yoneda lemma asserts that natural transformations from the functor represented by $\tilde Y$ to $\operatorname{Part}(-,Y)$ are in bijection with elements of $\operatorname{Part}(\tilde Y,Y)$, i.e. partial morphisms $\tilde Y\hookleftarrow U_Y\to Y$.

Moreover, the Yoneda lemma asserts the natural transformation corresponding to a fixed partial morphism $\tilde Y\overset{\eta_Y}\hookleftarrow U_Y\overset{f_Y}\to Y$ sends each morphism $\bar f\colon X\to\tilde Y$ to the image of the fixed partial morphism $(m_Y,f_Y)$ under the action $\operatorname{Part}(\bar f,Y)$ of the functor, i.e. to the partial morphism $(\bar f^*\eta_Y,f_Y\circ \eta_Y^{**}\bar f)$ as in the diagram: $$\require{AMScd} \begin{CD} U @>{\eta_Y^{**}\bar f}>> U_Y @>{f_Y}>> Y\\ @V{\bar f^*\eta_Y}VV @VVV \eta_Y \\ X @>{\bar f}>>\bar{Y} \end{CD}$$.

Representability of $\operatorname{Part}(-,Y)$ then means that the above natural transformation is bijective, i.e. that there is a bijective correspondence between partial morphisms $X\overset m\leftarrow U\overset f\to Y$ and morphisms $\bar f\colon X\to Y$ for which $(m,f)\cong(\bar f^*\eta_Y,\eta_Y^{**}\bar f)$.

The usual definition of partial morphism classifier a priori requires more than representability: it requires that the morphism $f_Y\colon U_Y\to Y$ of the partial morphism $(\eta_Y,f_Y)$ corresponding to the natural isomorphism of the functor represented by $\tilde Y$ with $\operatorname{Part}(-,Y)$ is the identitiy morphism.

Equivalence of the two definitions would then follow only if representability implies that $f_Y\colon U_Y\to Y$ is an isomorphism. Indeed, then $(\eta_Y,f_Y)$ and $(\eta_Y\circ f_Y^{-1},\mathrm{id}_Y)$ would be equivalant partial morphisms. That this is indeed the case as can be shown by checking the $f_Y$ has to be both a monomorphism and a split epimorphism.

First I claim that $f_Y\colon U_Y\to\tilde Y$ is a monomorphism, or more precisely, that a partial morphism $(\eta_Y,f_Y)$ corresponds to an injective natural transformation into $\operatorname{Part}(-,Y)$ only if $f_Y$ is a monomorphism.

To see this, note that the square in the diagram $$\require{AMScd} \begin{CD} X @>{g}>> U_Y @>{f_Y}>> Y\\ @V{\mathrm{id}_X}VV @VVV \eta_Y \\ X @>{\eta_Y\circ g}>>\bar{Y} \end{CD}$$ is always a pullback square. Thus, morphisms of the form $f_Y\circ g$ are in bijection not only with partial morphisms of the form $(\mathrm{id}_X,f_Y\circ g)$, but also with morphisms of the form $\eta_Y\circ g\colon X\to\tilde Y$ (by injectivity of the natural transformation), and then also with morphisms $g\colon X\to U_Y$ by $\eta_Y\colon U_Y\to\tilde Y$ being a monomorphism. But this means precisely that $f_Y\colon U_Y\to Y$ is a monomorphism.

Second, I claim that $f_Y\colon U_Y\to\tilde Y$ is a split epimorphism, i.e. that there exists $g\colon Y\to U_Y$ such that $f_Y\circ h=\mathrm{id}_Y$. Indeed, such would exist as a consequence of the surjectivity of the natural transformation onto the partial morphism $(\mathrm{id}_X,\mathrm{id}_X)$ which would have to arise as $$\begin{CD} Y @>{h}>> U_Y @>{f_Y}>> Y\\ @V{\mathrm{id}_Y}VV @VVV \eta_Y \\ Y @>{\bar{\mathrm{id}_Y}}>>\bar{Y} \end{CD}$$

It now follows that $f_Y\circ\mathrm{id}_Y=f_Y\circ h\circ f_Y$, whence $h\circ f_Y=\mathrm{id}_Y$ in addition to $f_Y\circ h=\mathrm{id}_Y$, i.e. $f$ is an isomorphism as desired.

To summarize: a partial morphism $(\eta_Y,f_Y)$ corresponds to an injective natural transformation into $\operatorname{Part}(-,Y)$ only if $f_Y$ is a monomorphism. Moreover, the natural transformation is surjective onto the total identity morphism only if $f_Y$ is a split epimorphism.

More abstractly: subfunctors of $\operatorname{Part}(-,Y)$ can only be represented by partial monomorphisms, and such a representable subfunctor includes the identity morphism (in which case also $\operatorname{Hom}(-,Y)$) if and only if is represented by a partial isomorphism, i.e. if it can be represented by a partial identity, i.e. single monomorphism.