Proving Segal's Category $\Gamma \simeq \mathbf{FinSet}_*^{op}$

Question

I'm trying to show that Segal's category $\Gamma$ is equivalent to $\mathbf{FinSet}_*^{op}$, the opposite of the category of finite pointed sets with basepoint preserving morphisms. I'm intuitively gathering that $\Gamma$ can be shown to be the skeleton of $\mathbf{FinSet}_*^{op}$, or equivalently that $\Gamma^{op}$ is the skeleton of $\mathbf{FinSet}_*$, but I'm having trouble just understanding what the opposites of these categories are.

For reference $\Gamma$ is the category consisting of finite sets, and morphisms from $S$ to $T$ are maps $\theta:S \to P(T)$ where $P(T)$ is the power set of $T$, such that distinct points in $S$ get mapped to disjoint subsets of $T$. If $\theta:S\to P(T)$ and $\varphi:T \to P(U)$ are two such morphisms, then their composite is given by a morphism $\psi: S \to P(U)$ such that

$$ \psi(\alpha) \;\; =\;\; \bigcup_{\beta \in \theta(\alpha)} \varphi(\beta). $$

My biggest issue here is that I can't even conceive of the opposite morphisms. How can these be found? If I wanted to demonstrate $\Gamma^{op} \simeq \mathbf{FinSet}_*$ and then use duality, I'd have to find the opposite morphisms $\theta^{op}$, but would these be mappings from $P(T)$ to $S$, or from $T$ to $P(S)$? What's even the form of the identity morphisms in $\Gamma$? I thought possibly these would be the mappings $1_S:S\to P(S)$ such that $1_S(\alpha) = \{\alpha\}$. In what sense can $1_S$ be seen as an isomorphism in $\Gamma$?

In what sense could we naturally find a notion of "opposite" here? For instance, for $A \in P(T)$ I'm doubtful we could define $\theta^{op}(A) = \theta^{-1}(A)$ since $\theta$ was a set-valued function to begin with. Either way, there's no guarantee that $A\in P(T)$ would be in the image of $\theta$.

Answer 1

Your typical morphism from $S$ to $T$ is a map $S\to\wp(T)$ satisfying the condition you list. This induces a map $\phi$ from $T_*$ to $S_*$ as follows, where $S_*$ is the disjoint union of $S$ and a basepoint $*$.

This map $\phi$ is defined by $\phi(t)=s$ whenever $t\in\theta(s)$ for some $s$ (necessarily unique as the $\theta(s)$ are disjoint, and $\phi(t)=*$ if $t=*$ or $t$ is not in any of the $\theta(s)$.

Then $S\mapsto S_*$ and $\theta\mapsto\phi$ are the object and morphism maps of the required contravariant functor.

Answer 2

I think it’s best to understand the equivalence $Γ ≃ \mathbf{FinSet}_*^{\mathrm{op}}$ via $$ Γ ≅ \mathbf{FinSet}_∂^{\mathrm{op}} ≃ \mathbf{FinSet}_*^{\mathrm{op}} \,, $$ where the intermediary term $\mathbf{FinSet}_∂$ is the category of finite sets and partially defined functions.

Partially defined functions

Recall that a partially defined function, or simply partial function, from a set $S$ to a set $T$ is a pair $(D, f)$ consisting of a subset $D$ of $S$ (the domain of the partial function) and a function $f \colon D \to T$. The composite of two partial functions $(D, f) \colon S \to T$ and $(E, g) \colon T \to U$ is the partial function $(F, h)$ with $F = D ∩ f^{-1}(E)$ and $h(s) = g(f(s))$ for every $s ∈ F$. This gives us a category $\mathbf{Set}_∂$ of sets and partial functions between them. We have a full subcategory $\mathbf{FinSet}_∂$ of $\mathbf{Set}_∂$ whose objects are the finite sets.

The isomorphism $Γ ≅ \mathbf{FinSet}_∂^{\mathrm{op}}$

If $(D, f)$ is a partial function from a set $S$ to a set $T$, then we can consider for every element $t$ of $T$ its fibre $f^{-1}(t)$. We get in this way a map $θ \colon T \to P(S)$ given by $θ(t) = f^{-1}(t)$ for every $t ∈ T$, and the sets $θ(t)$ and $θ(t')$ are disjoint for every two distinct element $t$ and $t'$ of $T$.

Conversely, if $θ \colon T \to P(S)$ is a map such that the sets $θ(t)$ and $θ(t')$ are disjoint for any two distinct elements $t$ and $t'$ of $T$, then we get a corresponding partial function $(D, f)$ from $S$ to $T$ with domain $D = \bigcup_{t ∈ T} θ(t)$ and mapping $f$ given by $f(s) = t$ for every $t ∈ T$ and every $s ∈ θ(t)$.

The above two constructions are mutually inverse, and result in an isomorphism (and not just an equivalence) of categories between $Γ$ and $\mathrm{FinSet}_∂^{\mathrm{op}}$.

The equivalences $\mathbf{Set}_∂ ≃ \mathbf{Set}_*$ and $\mathbf{FinSet}_∂ ≃ \mathbf{FinSet}_*$

Every partial function can be extended to a total function. To every set $S$ we add a new value named $\mathtt{undefined}$, resulting in a set $S_*$. Every partial function $(D, f) \colon S \to T$ can then extended to the total function $$ S_* \longrightarrow T_* \,, \quad s \longmapsto \begin{cases} f(s) & \text{if $s ∈ D$,} \\ \mathtt{undefined} & \text{otherwise.} \end{cases} $$ This extension maps $\mathtt{undefined}$ in $S_*$ to $\mathtt{undefined}$ in $T_*$, and we get overall a functor $F \colon \mathbf{Set}_∂ \to \mathbf{Set}_*$.

We have similarly a functor $G \colon \mathbf{Set}_* \to \mathbf{Set}_∂$ that removes from every pointed set $(S, s_0)$ its base-point $s_0$, resulting in the set $S ∖ \{ s_0 \}$, while restricting every pointed function $f \colon (S, s_0) \to (T, t_0)$ to a partial function $S ∖ \{s_0\} \to T ∖ \{ t_0 \}$ with domain $S ∖ f^{-1}(t_0)$.

The composite $G ∘ F$ is the identity functor on $\mathsf{Set}_*$. The composite $F ∘ G$ only renames the base point and is therefore still isomorphic to the identity functor on $\mathbf{Set}_*$ (the pointed set $(S, s_0)$ is mapped to the pointed set $(S ∖ \{s_0\} ∪ \{\mathtt{undefined}\}, \mathtt{undefined})$, which is not set-theoretically equal to $(S, s_0)$ but still isomorphic to it.)

We have a (fairly explicit) equivalence of categories $$ \mathbf{Set}_∂ ≃ \mathbf{Set}_* \,. $$ By restriction, we have also an equivalence of categories $$ \mathbf{FinSet}_∂ ≃ \mathbf{FinSet}_* \,. $$