I found the following problem in my lecture notes.
We need to prove $g^{-1}g$ is homotopic to $e_a$ where $g$ is a path from $a$ to $b$ and $e_a$ is the identity path at $a$. My lecture notes say that the homotopy is given by $F=g((1-t)q(s))$ when $0< s<1$ and $g^{-1}((1-t)q(s))$ when $1< s<2$ where $q(s)$ is given by $s$ for $0< s<1$ and $2-s$ for $1< s<2$.
The problem is that I think $q(s)$ should still be $s$ for $1< s<2$ since we want $g^{-1}$ to start at $b$ and end at $a$ when $t=0$.
Am I right or are the lecture notes corrrect as they are.
Thank you in advance and sorry for the terrible notation
Let $e_a:I\to X;\;e_a(s)=a,\;\forall s\in I$ be the constant loop.
Let $g:I\to X$ a path such that $g(0)=a$ and $g(1)=b$.
By definition, the inverse path $g^{-1}$ of $g$ is the path $$g^{-1}:I\to X; \; g^{-1}(s)=g(1-s), \;\forall s\in I$$ In particular this defines $g^{-1}$ as a path starting in $b$ and ending in $a$ and taking the same way as the path $g$ but in the opposite direction.
Also by definition, the product of the paths $g$ and $g^{-1}$ is $$gg^{-1}(s) = \begin{cases} g(2s) &\mbox{if }0\leq s\leq \dfrac{1}{2} \\ g^{-1}(2s-1)=g(2-2s)& \mbox{if } \dfrac{1}{2}\leq s\leq 1. \end{cases} $$ Now construct the homotopy $H_t:I\to X$ between $gg^{-1}$ and $e_a$ as follows: $$H_t(s)= \begin{cases} g((1-t)2s) &\mbox{if }0\leq s\leq \dfrac{1}{2} \\ g((1-t)(2-2s))& \mbox{if } \dfrac{1}{2}\leq s\leq 1. \end{cases} $$ Indeed, clearly $H_0=gg^{-1}$ and $\forall s\in I$, $H_1(s)=g(0)=a=e_a(s)$ hence $H_1=e_a$.