## Franklin's Notes

### Definition of a category

A category $\mathsf{C}$ consists of objects denoted $X,Y,Z,...$ and morphisms $f,g,h,...$ subject to the following rules:

1. Each morphism $f$ has a domain $X$ and a codomain $Y$, which can be expressed using the notation $f:X\to Y$.
2. Each object $X$ has an identity morphism $1_X:X\to X$.
3. For any pair of morphisms $f:X\to Y$ and $g:Y\to Z$, there exists a composite morphism $gf:X\to Z$.
4. For any $f:X\to Y$, we have $f1_X = 1_Yf = f$.
5. Morphism composition is associative. That is, for any morphisms $f:W\to X$, $g:X\to Y$, $h:Y\to Z$, we have $h(gf)=(hg)f$.

A subcategory is defined by restricting $\mathsf{C}$ to a subcollection of objects and morphisms such that the subcategory contains the domain and codomain of any selected morphism, the identity morphism of any object is included, and the composite of any pair of included morphisms is also included. If $\mathsf{C}$ is small, containing only a set's worth of morphisms, then a subcategory can be described as a subset which is contains the domain and codomain of each of its morphisms, and which is contains identities and is closed under morphism composition.

Riehl notes that, in some sense, the presence of objects in the definition of a category is superfluous. For objects can be identified in some sense with their identity maps, so that a category could be alternatively described as a set of morphisms including some two-sided identitiy morphisms and restrictions placed upon how composition is defined.

Exercise 1. (Riehl, 1.1.i)
1. Show that a morphism can have at most one inverse isomorphism.
2. Consider a morphism $f:x\to y$. Show that if there exists a pair of morphisms $g,h:y\rightrightarrows x$ so that $gf=1_x$ and $fh=1_y$, then $g=h$ and $f$ is an isomorphism.

Proof. For $(1)$, suppose that $f:x\to y$ and $g_1,g_2:y\rightrightarrows x$ are isomorphisms and inverses of $f$. Then $g_1 f = g_2 f = 1_x$ and $fg_1 = fg_2 = 1_y$. This means that and hence $g_1=g_2$ as desired.

For $(2)$, let $f,g,h$ be as specified in the question. If $gf=1_x$ and $fh=1_y$, then we have and $g=h$ as desired. $\blacksquare$