NOTE: slice category

A slice category $$A/a$$ is category, from a category $$A$$ and its object $$a$$, that follows

1. slice category has objects $$(b, f)$$, where
1. $$b$$ is $$A$$'s object.
2. and $$f$$ is $$A$$'s morphism.
2. 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.

Date: 2023-02-24 Fri 00:00