Category of Pointed Sets as Coslice Category

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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

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