# NOTE: slice category

A slice category \(A/a\) is category, from a category \(A\) and its object \(a\), that follows

- slice category has objects \((b, f)\), where
- \(b\) is \(A\)'s object.
- and \(f\) is \(A\)'s morphism.

- and has morphism \(g\), such that \(g\) is \(A\)'s morphism, which makes the following diagram commute.

As you see, the definitions above were picking objects and morphisms out to form a new category. Therefore, the above diagram made a corresponding \(A/a\) diagram.

## 1. \((a, id_a)\) is \(A/a\)'s terminal object

It's easy to prove the statement by checking the following diagram is commute.

or we can also reduce the diagram to

Draw a \(A/a\) diagram makes it clear, the following relationship stands for every proper \(b\) in \(A\).

## 2. Any \(f : a \to a'\) in \(A\) gives a functor \(f_* : A/a \to A/a'\)

The functor \(f_*\) takes any object \(g : b \to a\) of \(A/a\) to \(f \circ g\) of \(A/a'\)

The first point is realizing if such \(f\) existed, then we have same two \(b\), \(b'\) for both \(a\) and \(a'\), by concating \(f\) after every \(b \to a\) morphism. This makes sure we will have if objects existed in \(A/a\) so do in \(A/a'\), remember functor no need to cover \(A/a'\). Now, we can have the proof diagram:

Convert it to a functor diagram will help.