Is known that the slice category $\mathbb{Set}/I$ is equivalent to the category of $I$-indexed sets $\mathbb{Set}^{I}$. We can establish two functors
$$\varphi: \mathbb{Set}^{I} \rightarrow \mathbb{Set}/I$$ $$\psi: \mathbb{Set}/I \rightarrow \mathbb{Set}^{I}$$
by
- $\varphi \left( \left( A_{i} \right)_{i \in I} \right) \mapsto \pi_{A}: \oplus_{i \in I} A_{i} \rightarrow I $, where $\oplus$ stands for the disjoint union and $\pi$ is the indexing projection, and
- $\psi \left( \pi: A \rightarrow I \right) \mapsto \oplus_{i \in I} \{ \pi^{-1}(i) \} $.
It is then easy to show the equivalence $ \mathbb{1}_{ \mathbb{Set}/I } \simeq \varphi \psi, \mathbb{1}_{ \mathbb{Set}^{I} } \simeq \psi \varphi $.
But I cannot see how these functors are not essentially the inverses of each other. The operations seem to be completely reversible and we get exactly the original result after applying $\psi \varphi$ or $\varphi \psi$.
I could not come up with a counterexample. Any help is appreciated.
Recall that the disjoint union of sets is usually constructed as $\coprod_i A_i = \bigcup A_i \times \{i\}$, but other constructions are also available. If $A$ is a set with a map $A \to I$ and fibers $A_i$ for $i \in I$, then $A$ is not necessarily equal to that disjoint union $\coprod_i A_i$; rather we have a canonical isomorphism $\coprod_i A_i \to A$. This shows that $\phi \psi$ is isomorphic, but not equal to the identity. Probably we can remedy this by constructing $\coprod_i A_i$ as $\bigcup_i A_i$ if the sets $A_i$ are pairwise disjoint, and otherwise as $\bigcup_i A_i \times \{i\}$. But still, in the other direction, if we have a family of sets $(A_i)_{i \in I}$, then the fiber of $\coprod_i A_i \to I$ will be $A_i \times \{i\}$ and only equals $A_i$ if the given family of sets is pairwise disjoint. As you can see, these details really belong to set theory and have no "categorical content". See also principle of equivalence.