Category of Pointed Sets as Coslice Category
Theorem
Let $\mathbf{Set}_*$ be the category of pointed sets.
Let $\mathbf{Set}$ be the category of sets.
Let $1 := \left\{{*}\right\}$ be any singleton.
Then:
- $\mathbf{Set}_* \cong 1 \mathbin / \mathbf{Set}$
where $1 \mathbin / \mathbf{Set}$ denotes the coslice of $\mathbf{Set}$ under $1$ and $\cong$ signifies isomorphic categories.
Proof
Define the functor $F: \mathbf{Set}_* \to 1 \mathbin / \mathbf{Set}$ by:
- $\map F {C, c} := \bar c: 1 \to C$
- $F f := f$
where $\bar c: 1 \to C$ is defined by $\bar c (*) = c$.
Further, define $G: 1 \mathbin / \mathbf{Set} \to \mathbf{Set}_*$ by:
- $\map G {x: 1 \to C} := \struct {C, x \paren *}$
- $G f := f$
$F$ is a functor
The definition of $F$ on morphisms is admissible, since for any pointed mapping $\map f: {C, c} \to \struct {D, d}$:
\(\ds f \circ \bar c \paren *\) | \(=\) | \(\ds \map f c\) | Definition of $\bar c$ | |||||||||||
\(\ds \) | \(=\) | \(\ds d\) | $f$ is a pointed mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \bar d \paren *\) | Definition of $\bar d$ |
Thus by Equality of Mappings, $f \circ \bar c = \bar d$.
So indeed $F f = f$ is a morphism $\bar c \to \bar d$, as desired.
The identity morphisms of both $\mathbf{Set}_*$ and $1 \mathbin / \mathbf{Set}$ are the identity mappings, which $F$ thus preserves.
That $F$ preserves composition is also trivial, since $\mathbf{Set}_*$ and $1 \mathbin / \mathbf{Set}$ both have composition of mappings as their $\circ$.
In conclusion, $F$ is a functor.
$G$ is a functor
Let $x: 1 \to C$ and $y: 1 \to D$ be objects of the coslice $1 \mathbin / \mathbf{Set}$.
Let $f: C \to D$ be a morphism $x \to y$.
That is, let $f \circ x = y$.
Thus, in particular:
- $\map f {\map x *} = \map y *$
showing that $f$ is a pointed mapping $\struct {C, \map x *} \to \struct {D, \map y *}$.
Observe that the composition and the identity morphisms of $1 \mathbin / \mathbf{Set}$ and $\mathbf{Set}_*$ are identical.
Because $G$ is the identity on morphisms, it is thus trivially a functor.
$F$ is an isomorphism
Because $F$ and $G$ are the identity on morphisms, it will suffice to show that:
- $F \map G x = x$ for all objects $x$ of $1 \mathbin / \mathbf{Set}$
- $G \map F {C, c} = \struct {C, c}$ for all objects $\struct {C, c}$ of $\mathbf{Set}_*$
Explicitly:
\(\ds F \map G {x: 1 \to C}\) | \(=\) | \(\ds \map F {C, \map x *}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \overline {\map x *}\) |
Now $\overline {\map x *}: 1 \to C$ is defined by $\map {\overline {\map x *} * = \map x *$.
Hence $\overline {\map x *} = x$ by Equality of Mappings.
For the other equality:
\(\ds G \map F {C, c}\) | \(=\) | \(\ds \map G {\bar c: 1 \to C}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \struct {C, \map {\bar c} *}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \struct {C, c}\) |
where the last equality follows by definition of $\bar c$.
Thus $F$ is shown to be an isomorphism, and hence:
- $\mathbf{Set}_* \cong 1 \mathbin / \mathbf{Set}$
$\blacksquare$
Sources
- 2010: Steve Awodey: Category Theory (2nd ed.) ... (previous) ... (next): $\S 1.6$: Example $1.8$