Category

A category

𝒞

consists of

a collection ob $(𝒞)$ of objects
for each $A, B \in ob (𝒞)$ , a collection $mor (A, B)$ of what we call morphisms / arrows from $A$ to $B$
for each $A, B, C \in ob (𝒞)$ , a function which we call composition $\circ_{𝒞} : mor (B, C) \times mor (A, B) \to mor (A, C)$

\circ_{𝒞} (g, f) : = g \circ f

for each $A \in ob (𝒞)$ , an element $1_{A}$ of $mor (A, A)$ , called the identity on $A$ , satisfying the following axioms:

associativity: for each $f \in mor (A, B), g \in mor (B, C)$ and $h \in mor (C, D)$ , we have $(h \circ g) \circ f = h \circ (g \circ f)$ ;
identity laws: for each $f \in mor (A, B)$ , we have $f \circ 1_{A} = f = 1_{B} \circ f$ .

Domain and Codomain

Each morphism has associated domain and codomain, denoted

f : A \to B

The domain of $f$ , denoted $dom (f)$ , is $A$
The codomain of $f$ , denoted $cod (f)$ , is $B$

Some Notations on Morphism

$f, g : A ⇉ B$ denotes two morphisms with the same domain and codomain
$f : A ⇄ B : g$ denotes two morphisms, $f : A \to B$ and $g : B \to A$

Isomorphism / Isomorphic

Let

𝒞

be a category,

A, B \in ob (𝒞)

f : A \to B

is an isomorphism if and only if

\begin{matrix} \exists g : B \to A . & g \circ f = 1_{A} \land \\ f \circ g = 1_{B} \end{matrix}

In which case

A, B

are isomorphic and denoted as

A ≅ B

Endomorphism

Let

𝒞

be a category,

A \in ob (𝒞)

f : A \to A

is an endomorphism if and only if

f

has the same domain and codomain

Automorphism

A morphism

f

is an automorphism if and only if

f is an endomorphism \land f is an isomorphism

Opposite Category, Duality

Let

𝒞

be a category. Then the opposite category

𝒞^{op}

is defined as follow:

$ob (𝒞^{op}) = ob (𝒞)$
For each $f \in {mor}_{𝒞} (B, A)$ , a morphism $f^{op} \in {mor}_{𝒞^{op}} (A, B)$
$g^{op} \circ_{𝒞^{op}} f^{op} = {(f \circ_{𝒞} g)}^{op}$

The opposite category is a category

The construction

𝒞^{op}

indeed form a new category.

The definition already gives us the mathematical objects of a category. We only need to show composition satisfies associativity and identity law.

We first show associativity. Let $f^{op} : A \to B, g^{op} : B \to C, h^{op} : C \to D$ be morphisms in $𝒞^{op}$ . Correspondingly, we have morphisms $f : B \to A, g : C \to B, h : D \to C$ in $𝒞$ . We then have the following reasoning (composition subscript omitted): $\begin{matrix} (h^{op} \circ g^{op}) \circ f^{op} \\ = & {(g \circ h)}^{op} \circ f^{op} \\ = & {(f \circ (g \circ h))}^{op} \\ = & {((f \circ g) \circ h)}^{op} \\ = & h^{op} \circ {(f \circ g)}^{op} \\ = & h^{op} \circ (g^{op} \circ f^{op}) \end{matrix}$

Lastly we show the existence of identity morphism along with their desired property. Let $A, B \in 𝒞^{op}$ , we choose the identity morphism of this object to be $1_{A}^{op}, 1_{B}^{op}$ respectively.

Let $f^{op} : A \to B$ be a morphism in $𝒞$ . We can show the following: $\begin{matrix} f^{op} \circ 1_{A}^{op} = & {(1_{A} \circ f)}^{op} \\ = & {(f)}^{op} \\ 1_{B}^{op} \circ f^{op} = & {(f \circ 1_{B})}^{op} \\ = & {(f)}^{op} \end{matrix}$

Thus all desired properties are shown, we conclude $𝒞^{op}$ is a category.

Functor

Let

𝒜

and

ℬ

be categories. A functor

F : 𝒜 \to ℬ

consists of

a function $ob (𝒜) \to ob (ℬ)$ , written as $A \mapsto F (A)$
- for each $A, A^{'} \in 𝒜$ , a function ${mor}_{𝒜} (A, A^{'}) \to {mor}_{ℬ} (F (A), F (A^{'}))$ written as $f \mapsto F (f)$ , satisfying the following axioms:

$F (f^{'} \circ f) = F (f^{'}) \circ F (f)$ whenever $A \overset{\overset{}{f}}{\to} A^{'} \overset{\overset{}{f^{'}}}{\to} A^{''}$ in $𝒜$ ;
$F (1_{A}) = 1_{F (A)}$ whenever $A \in 𝒜$ .

Pointed Topological Spaces Category

The category of pointed topological spaces denoted by

{Top}_{0}

is one where objects are pairs

(X, x_{0})

, where

X

is a topological space, and

x_{0} \in X

is a point of

X

. Morphisms between

(X, x_{0})

and

(Y, y_{0})

are continuous functions

f : X \to Y

such that

f (x_{0}) = y_{0}

Group Category

The category

Grp

has the class of all groups for objects and group homomorphisms for morphisms

The Fundamental Group Is a Functor

π_{1} : {Top}_{0} \to Grp

can be completed to a functor

For each $(X, x_{0}) \in ob ({Top}_{0})$ we map it to $π_{1} (X, x_{0})$ which is the object functor. We call this function $π_{1}^{ob} : ob ({Top}_{0}) \to ob (Grp)$

Now we will construct our morphism functor, that is given $f \in hom ((X, x_{0}), (Y, y_{0}))$ we map it to a function , defined by $[γ] \mapsto [f \circ γ]$ we call this mapping $h$ , that is: $F_{(X, x_{0}), (Y, y_{0})}$ where $F_{(X, x_{0}), (Y, y_{0})} (f) = h$

We now show that $h \in hom (π_{1} (X, x_{0}), π_{1} (Y, y_{0}))$ , to do this, we must prove that it is a group homomorphism, and also that it is well defined. For the well definedness we note that if $[α] = [β]$ then we have that $[f \circ α] = [f \circ β]$ . For ease of notation recall that $f_{*} ([γ]) : = [f \circ γ]$ , and we prove that our function $h$ is a group homomorphism, that is we show that $f_{*} ([α] [β]) = f_{*} ([α]) f_{*} ([β])$ but on the left we have $\begin{matrix} f_{*} ([α] \cdot [β]) & = f_{*} ([α \cdot β]) \\ = [f \circ (α \cdot β)] \\ = [(f \circ α) \cdot (f \circ β)] \\ = [f \circ α] \cdot [f \circ β] \\ = f_{*} (α) \cdot f_{*} (β) \end{matrix}$ which is the right, so we are done.

We now verify the functor properties: $F_{(X, x_{0}), (X, x_{0})} ({id}_{(X, x_{0})}) = {id}_{π_{1}^{ob} (X, x_{0})} = {id}_{π_{1} (X, x_{0})}$ We have that $F_{(X, x_{0}), (X, x_{0})} ({id}_{(X, x_{0})}) = h$ where $h ([γ]) = [{id}_{(X, x_{0})} \circ γ] = [γ]$ , in other words $h$ takes an element from $π_{1} (X, x_{0})$ and returns that same element, so that $h = {id}_{π_{1} (X, x_{0})}$ as needed.

We now verify that $F_{(X, x_{0}), (Y, y_{0})} (g \circ f) = F_{(Y, y_{0}), (Z, z_{0})} (g) \circ F_{(X, x_{0}), (Y, y_{0})} (f)$ If we focus on the right hand side and give the two functions simpler names ; $h_{g} : π_{1} (Y, y_{0}) \to π_{1} (Z, z_{0})$ and $h_{f} : π_{1} (X, x_{0}) \to π_{1} (Y, y_{0})$ respectively, and thus on the right hand side we have: $\begin{matrix} (h_{g} \circ h_{f}) ([γ]) & = h_{g} (h_{f} ([γ])) \\ = h_{g} ([f \circ γ]) \\ = [g \circ (f \circ γ)] \\ = [(g \circ f) \circ γ] \\ = F_{(X, x_{0}), (Y, y_{0})} (g \circ f) \end{matrix}$ so our original equation holds true.

Thus since we've verified all the conditions, we conclude that $π_{1}$ is a functor.