For sets and functions, I think the following data are equivalent:


1. A function $g:A\times B\to B$ such that $(\pi\_1,g):A\times B\to A\times B$ is a bijection;
2. a function $A\to \mathrm{Aut}B$.


*Proof.* The first condition is equivalent to $(\pi\_1,g)$ having singleton fibers. The fiber of $a,b$ is the intersection $(\left\{ a \right\} \times B)\cap g^{-1} \left\{ b \right\}$, which is exactly the fiber of $g|\_{ \left\{ a \right\} \times B}$. Thus $(\pi\_1,g)$ is bijective iff all of the restrictions are. These restrictions are precisely the same as a function $A\to \mathrm{Aut}B$. $\square$


Motivated by the transition functions of covering maps, I am hoping for the following statement.


**Tentative proposition.** Let $\Pi\_0:\mathsf{Fam}(\mathsf A)\to \mathsf{Set}$ be a fibration of connected components and let $H$ be its right adjoint, the discrete functor. Then the following data are equivalent:


1. An automorphism over $X$ of the projection $X\times HA\to X$;
2. An arrow $g:X\times HA\to HA$ such that $(\pi\_1,g):X\times HA\to X\times HA$ is an isomorphism;
3. An arrow $X\to H\mathrm{Aut}A$.


The first two conditions are equivalent in any category, but I'm having trouble exporting of the equivalence of the final two conditions from the category of sets. The distributivity of $\mathsf{Fam}(\mathsf A)$ and $\mathsf{Set}$ along with the adjunction $\Pi\_0\dashv H$ and the Cartesian closedness $\mathsf{Set}$ yield natural isomorphisms $$\mathsf{Fam}(\mathsf A)(X\times HA,HA)\cong \mathsf{Fam}(\mathsf A)(X,H\mathrm{End}A),$$ but I also don't know how to get the statement from here.


Is the tentative proposition correct? How can I prove it?



General description of transition arrows of covering morphisms in family fibrations
I believe this is true. Note that $\mathrm{Aut}(A)$ is a triple equalizer of three maps $\mathrm{End}(A) \times \mathrm{End}(A) \to \mathrm{End}(A)$, the composites in both orders and the constant map at $\mathrm{id}\_A$. Since $H$ is a right adjoint, it preserves limits, and in particular this triple equalizer. Moreover, $\mathrm{End}(A)$ is itself a limit, namely the $A$-fold power of $A$, so $H \mathrm{End}(A)$ is the $A$-fold power of $H A$, which is also the exponential $(H A)^{H A}$ by distributivity of $\mathsf{Fam(A)}$. Thus $H \mathrm{Aut}(A)$ is the triple equalizer of the corresponding three maps $(H A)^{H A} \times (H A)^{H A} \to (H A)^{H A}$, making it $\mathrm{Aut}(H A)$ (even if we don't assume that $\mathsf{Fam(A)}$ has all limits, it has this one). Therefore, a map $X \to H \mathrm{Aut}(A)$ is the same as a map $X \to \mathrm{Aut}(H A)$.


The rest of the argument doesn't need $H$ at all, so let $Y = H A$, which might as well be any object of any category such that the triple equalizer $\mathrm{Aut}(Y)$ of the three maps $Y^Y \times Y^Y \to Y^Y$ exists; call these three maps $\mathrm{comp}$, $\mathrm{comp}'$, and $\mathrm{const}\_{\mathrm{id}}$. By the universal property of this triple equalizer, a map $X\to \mathrm{Aut}(Y)$ consists of two maps $\ulcorner f \urcorner X\to Y^Y$ and $\ulcorner g \urcorner X\to Y^Y$ such that $\mathrm{comp}(\ulcorner f \urcorner,\ulcorner g \urcorner) = \mathrm{const\_{\mathrm{id}}} = \mathrm{comp}(\ulcorner g \urcorner,\ulcorner f \urcorner)$ (since $\mathrm{comp}(a,b) = \mathrm{comp}(b,a)$ by definition). By the exponential adjunction $\ulcorner f \urcorner$ and $\ulcorner g \urcorner$ are determined by maps $f:X\times Y \to Y$ and $g:X\times Y\to Y$, and the above equalities translate to say that the composites $X\times Y \xrightarrow{(\mathrm{id},f)} X\times Y \xrightarrow{g} Y$ and $X\times Y \xrightarrow{(\mathrm{id},g)} X\times Y \xrightarrow{f} Y$ are the second projection, or equivalently that $(\mathrm{id},f)$ and $(\mathrm{id},g)$ are inverse isomorphisms, i.e. we have an automorphism of $X\times Y$ in the slice category over $X$, which is your condition 1.


