# Continuous functions between topological spaces and their homotopy equivalence relations

Let $A,B,C$ be topological spaces and $\alpha,\alpha':A\rightarrow B$ continuous and $\beta,\beta':B\rightarrow C$ be continuous. Let $\sim$ be the homotopy relation (which I know/can use to be an equivalence relation).
Further let $\alpha\sim\alpha'$ and $\beta\sim\beta'$.

1 How do I prove that $\beta\circ\alpha,\beta'\circ\alpha':A\rightarrow C$ also have $\beta\circ\alpha\sim\beta'\circ\alpha'$?
2 How do I show that there is a unique map $$H(B,C)\times H(A,B)\xrightarrow{*} H(A,C)$$ of sets such that for all continuous $f:A\rightarrow B$ and continuous $g:B\rightarrow C$ we get $$\langle g\rangle* \langle f\rangle=\langle g\circ f\rangle?$$ Here $H(A,B)$ is the homotopy class of continuous maps $A\rightarrow B$, analog for $H(B,C)$ and $\langle f\rangle$ is the class of $f$ in $H(A,B)$.

What I know:
A homotopy from $\alpha$ to $\alpha'$ is a continuous map $\xi:[0,1]\times A\rightarrow B$ such that for all $x\in A$ we get $$\xi(0,x)=\alpha(x)\text{ and }\xi(1,x)=\alpha'(x).$$ Furthermore being an equivalence relation means it is
1. reflexive
2. symmetric
3. transitive

For question 1, since $\alpha \sim \alpha'$ and $\beta \sim \beta'$, there exists:

1. $\xi : [0,1] \times A \to B$ such that $\xi(0,x) = \alpha(x)$ and $\xi(1,x) = \alpha'(x)$;
2. $\zeta : [0,1] \times B \to C$ such that $\zeta(0,y) = \beta(y)$ and $\zeta(1,y) = \beta'(y)$.

Now define $\theta : A \times [0,1] \to C$ by $$\theta(t,x) = \zeta(t,\xi(t,x)).$$ Then $$\theta(0,x) = \zeta(0,\xi(0,x)) = \beta(\alpha(x)) = (\beta \circ \alpha)(x),$$ $$\theta(1,x) = \zeta(1,\xi(1,x)) = \beta'(\alpha'(x)) = (\beta' \circ \alpha')(x),$$ so $\theta$ is a homotopy between $\beta \circ \alpha$ and $\beta' \circ \alpha'$. Hence $\beta \circ \alpha \sim \beta' \circ \alpha'$.

For question 2, the unicity is clear. Since $H(X,Y)$ is defined as a quotient of a set. Composition gives a map $$\circ : \operatorname{Map}(B,C) \times \operatorname{Map}(A,B) \to \operatorname{Map}(A,C).$$

Compose this with the projection $\operatorname{Map}(A,C) \to H(A,C)$ to get a map $$\circ' : \operatorname{Map}(B,C) \times \operatorname{Map}(A,B) \to H(A,C).$$ This map is given by $(\beta,\alpha) \mapsto \langle \beta \circ \alpha \rangle$. Since $\beta \sim \beta$ and $\alpha \sim \alpha$, you get, using the result of question 1:

1. If $\alpha \sim \alpha'$, then for all $\beta$, $\beta \circ \alpha \sim \beta \circ \alpha'$;
2. If $\beta \sim \beta'$, then for all $\alpha$, $\beta \circ \alpha \sim \beta' \circ \alpha$.

The first point shows, by the mapping property of quotient by an equivalence relation*, that $\circ'$ induces a map \begin{align} \operatorname{Map}(B,C) \times H(A,B) & \to H(A,C) \\ (\beta, \langle \alpha \rangle) & \mapsto \langle \beta \circ \alpha \rangle; \end{align} the second point shows that this map also induces a map \begin{align} H(B,C) \times H(A,B) & \to H(A,C) \\ (\langle \beta \rangle, \langle \alpha \rangle) & \mapsto \langle \beta \circ \alpha \rangle; \end{align}

* This is the property that says that if $A$, $B$ are sets and $\sim$ is an equivalence relation of $A$, then a map $f : A \to B$ satisfying $a \sim a' \implies f(a) = f(a')$ induces a unique map $A/{\sim} \to B$ given by $\langle a \rangle \mapsto f(a)$.