## Functors, Homotopy Sets and Groups

Functors
Definition:
A functor $F$ from a category $\mathscr{C}$ to a category $\mathscr{D}$ is a function which
– For each object $X\in\mathscr{C}$, we have an object $F(X)\in\mathscr{D}$.
– For each $f\in\hom_\mathscr{C}(X,Y)$, we have a morphism $\displaystyle F(f)\in\hom_\mathscr{D}(F(X),F(Y)).$

Furthermore, $F$ is required to satisfy the two axioms:
– For each object $X\in\mathscr{C}$, we have $F(1_X)=1_{F(X)}$. That is, $F$ maps the identity morphism on $X$ to the identity morphism on $F(X)$.

– For $f\in\hom_{\mathscr{C}}(X,Y)$, $g\in\hom_\mathscr{C}(Y,Z)$ we have $\displaystyle F(g\circ f)=F(g)\circ F(f)\in\hom_\mathscr{D}(F(X),F(Z)).$ That is, functors must preserve composition of morphisms.

Definition:
A cofunctor (also called contravariant functor) $F^*$ from a category $\mathscr{C}$ to a category $\mathscr{D}$ is a function which
– For each object $X\in\mathscr{C}$, we have an object $F^*(X)\in\mathscr{D}$.
– For each $f\in\hom_\mathscr{C}(X,Y)$ we have a morphism $\displaystyle F^*(f)\in\hom_\mathscr{D}(F^*(Y),F^*(X))$ satisfying the two axioms:
– For each object $X\in\mathscr{C}$ we have $F^*(1_X)=1_{F^*(X)}$. That is, $F^*$ preserves identity morphisms.
– For each $f\in\hom_\mathscr{C}(X,Y)$ and $g\in\hom_\mathscr{C}(Y,Z)$ we have $\displaystyle F^*(g\circ f)=F^*(f)\circ F^*(g)\in\hom_\mathscr{D}(F^*(Z),F^*(X)).$ Note that cofunctors reverse the direction of composition.

Example

Given a fixed pointed space $(K,k_0)\in\mathscr{PT}$, we define a functor $\displaystyle F_K:\mathscr{PT}\to\mathscr{PS}$ as follows: for each $(X,x_0)\in\mathscr{PT}$ we assign $F_K(X,x_0)=[K,k_0; X,x_0]\in\mathscr{PS}$. Given $f: (X,x_0)\to (Y,y_0)$ in $\hom((X,x_0),(Y,y_0))$ we define $F_K(f)\in\hom([K,k_0; X,x_0],[K,k_0;Y,y_0])$ by $\displaystyle F_k(f)[g]=[f\circ g]\in[K,k_0; Y,y_0]$ for every $[g]\in [K,k_0; X,x_0]$.

We can check the two axioms:
– $F_k(1_X)[g]=[1_X\circ g]=[g]$ for every $[g]\in[K,k_0; X, x_0]$.
– For $f\in\hom((X,x_0),(Y,y_0))$, $h\in\hom((Y,y_0),(Z,z_0))$ we have $\displaystyle F_K(h\circ f)[g]=[h\circ f\circ g]=F_K(h)\circ F_K(f)[g]\in[K,k_0; Z,z_0]$ for every $[g]\in[K,k_0; X,x_0]$.

Similarly, we can define a cofunctor $F_K^*$ by taking $F_K^*(X,x_0)=[X,x_0; K,k_0]$ and for $f:(X,x_0)\to (Y,y_0)$ in $\hom((X,x_0),(Y,y_0))$ we define $\displaystyle F_K(f)[g]=[g\circ f]\in[X,x_0; K,k_0]$ for every $[g]\in[Y,y_0; K,k_0]$.

Note that if $f\simeq f'$ rel $x_0$, then $F_K(f)=F_K(f')$ and similarly $F_K^*(f)=F_K^*(f')$. Therefore $F_K$ (resp.\ $F_K^*$) can also be regarded as defining a functor (resp.\ cofunctor) $\mathscr{PT}'\to\mathscr{PS}$.

Homotopy Sets and Groups
Theorem:
If $(X,x_0)$, $(Y,y_0)$, $(Z,z_0)\in\mathscr{PT}$, $X$, $Z$ Hausdorff and $Z$ locally compact, then there is a natural equivalence $\displaystyle A: [Z\wedge X, *; Y,y_0]\to [X, x_0; (Y,y_0)^{(Z,z_0)}, f_0]$ defined by $A[f]=[\hat{f}]$, where if $f:Z\wedge X\to Y$ is a map then $\hat{f}: X\to Y^Z$ is given by $(\hat{f}(x))(z)=f[z,x]$.

We need the following two propositions in order to prove the theorem.

Proposition 1:
The exponential function $E: Y^{Z\times X}\to (Y^Z)^X$ induces a continuous function $\displaystyle E: (Y,y_0)^{(Z\times X, Z\vee X)}\to ((Y,y_0)^{(Z,z_0)}, f_0)^{(X,x_0)}$ which is a homeomorphism if $Z$ and $X$ are Hausdorff and $Z$ is locally compact\footnote{every point of $Z$ has a compact neighborhood}.

Proposition 2:
If $\alpha$ is an equivalence relation on a topological space $X$ and $F:X\times I\to Y$ is a homotopy such that each stage $F_t$ factors through $X/\alpha$, i.e.\ $x\alpha x'\implies F_t(x)=F_t(x')$, then $F$ induces a homotopy $F':(X/\alpha)\times I\to Y$ such that $F'\circ (p_\alpha\times 1)=F$.