How can I prove "Homotopy is an equivalence relation on the set of all loops on $X$." using the definition of homotopy $rel$ for loops:
Let $f, g$ be two loops in $X$ based at $x$, i.e. $f(0) = f(1) = x = g(0) = g(1)$. We say that $f$ is homotopic to $g$ and write $f \sim{g}$ if $f \sim{g}$ $rel_A$ where $A=\{0,1\} $, i.e if there exists a homotopy $$H : I \times I \rightarrow X, $$ such that $$H(s,0) = f(s),H(s,1) = g(s), \forall s \in I,$$ and $$H(0,t) = H(1,t) = x, \forall t \in I.$$
I can prove it like in path. But here I didn't use the relative homotopy for loops neither the fact that $f(0)=f(1)=x$.
(i) Reflexivity ($f\sim f$). Let $X$ and $Y$ be two topological spaces and $f:X\longrightarrow Y$ be a loop. Define $F:X\times I\longrightarrow Y$ by $F(x,t)=f(x), \forall x\in X.$ Then $F$ is continuous and $F(x,0)=F(x,1)=f(x)$ for all $x\in X.$ So $f\sim f.$
(ii) Symmetry ($f\sim g\Rightarrow g\sim f$). Suppose that $f,g:X\longrightarrow Y$ are two loops and $f\sim g.$ Then there is a map $F: X\times I\longrightarrow Y$ such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$ for every $x\in X.$ Define a mapping $G:X\times I \longrightarrow Y$ by $G(x,t)=F(x,1-t).$ Then $G$ is continuous, $G(x,0)= F(x,1)=g(x)$ and $G(x,1)=F(x,0)=f(x)$, for every $x\in X.$ Therefore, $g\sim f.$
(iii) Transitivity ($f\sim g , g\sim h\Rightarrow f\sim h$). Suppose that $f,g,h:X\longrightarrow Y$ are three loops such that $f\sim g$ and $g\sim h.$ Then there is a map $F:X\times I\longrightarrow Y$ such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$ for every $x\in X$ and another map $G:X\times I\longrightarrow Y$ such that $G(x,0)=g(x)$ and $G(x,1)=h(x)$, for every $x\in X.$ Define a mapping $H:X\times I\longrightarrow Y$ by \begin{align*} H(x,t) = \begin{cases} F(x,2t)&, 0\leq t\leq\dfrac{1}{2}, \\ G(x,2t-1)&, \dfrac{1}{2}\leq t\leq 1. \end{cases} \end{align*} It is clear that $F$ is continuous in $[0,\dfrac{1}{2}]$ and $G$ is continuous in $[\dfrac{1}{2},1].$ Hence, according to glueing lemma $H$ is continuous on $[0,1].$ Moreover, $H(x,0)= F(x,0)=f(x)$ and $H(x,1)=G(x,1)=h(x).$ Thus $f \underset{H}{\thicksim} h.$
Therefore homotopy is an equivalence relation.