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:

$F \left({C, c}\right) := \bar c: 1 \to C$
$F f := f$

where $\bar c: 1 \to C$ is defined by $\bar c (*) = c$.

Further, define $G: 1 \mathbin / \mathbf{Set} \to \mathbf{Set}_*$ by:

$G \left({x: 1 \to C}\right) := \left({C, x (*)}\right)$
$G f := f$


$F$ is a functor

The definition of $F$ on morphisms is admissible, since for any pointed mapping $f: \left({C, c}\right) \to \left({D, d}\right)$:

\(\displaystyle f \circ \bar c (*)\) \(=\) \(\displaystyle f (c)\) Definition of $\bar c$
\(\displaystyle \) \(=\) \(\displaystyle d\) $f$ is a pointed mapping
\(\displaystyle \) \(=\) \(\displaystyle \bar d (*)\) 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:

$f \left({x (*)}\right) = y (*)$

showing that $f$ is a pointed mapping $\left({C, x(*)}\right) \to \left({D, y(*)}\right)$.

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:

$F G (x) = x$ for all objects $x$ of $1 \mathbin / \mathbf{Set}$
$G F \left({C, c}\right) = \left({C, c}\right)$ for all objects $\left({C, c}\right)$ of $\mathbf{Set}_*$

Explicitly:

\(\displaystyle F G \left({x: 1 \to C}\right)\) \(=\) \(\displaystyle F \left({C, x (*)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \overline{x (*)}\)

Now $\overline{x (*)}: 1 \to C$ is defined by $\overline{x (*)} (*) = x (*)$.

Hence $\overline{x (*)} = x$ by Equality of Mappings.

For the other equality:

\(\displaystyle G F \left({C, c}\right)\) \(=\) \(\displaystyle G \left({\bar c: 1 \to C}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({C, \bar c (*)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({C, c}\right)\)

where the last equality follows by definition of $\bar c$.


Thus $F$ is shown to be an isomorphism, and hence:

$\mathbf{Set}_* \cong 1 \mathbin / \mathbf{Set}$

$\blacksquare$


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