Homotopy with paths and inverse paths and a counterexample Exercise 3.6 Rotman

(i) If $f\simeq g$ rel $I^\circ$ (where $I^\circ$ means $\{0,1\}$, the boundary of $I$), then $f^{-1}\simeq g^{-1}$ rel $I^\circ$, where $f$ and $g$ are paths in $X$.
(ii) If $f$ and $g$ are paths in $X$ with $\omega(f)=\alpha(g)$ then $(f*g)^{-1}=g^{-1}*f^{-1}$.
(iii) Give an example of a closed path $f$ where $f*f^{-1}\neq f^{-1}*f$ .
(iv) Show that if $\alpha(f)=p$ and $f$ is not constant then $i_p*f\neq f$.

Attempts to each of them
(i) Consider $H:f\simeq g$ rel $I^\circ$ a relative homotopy. We want $G:f^{-1}\simeq g^{-1}$ rel $I^\circ.$
If we draw a "triangular" diagram from $I\times I\to X, I\times I\to I\times I, I\times I \to X$ under $G,(s,t)\mapsto(1-s,t),H$, respectively, we find that $G=H\circ(1-s,t).$ This indeed yields to $G:f^{-1}\simeq g^{-1}$ rel $I.$
(ii) I do not know how to write $(f*g)^{-1}$ explicitly to try to find similarities with the right hand side $g^{-1}*f^{-1}$.
(iii) How can I come up with a counterexample such that $(f*g)^{-1}=g^{-1}*f^{-1}$?
(iv) By contradiction.
Suppose $i_p*f= f.$
Thus $f(t)=i_p*f(t)=\begin{cases}i_p(2t),0\le t\le\frac 12\\ f(2t-1),\frac 12\le t,\le 1 \end{cases}$
If $t=\frac 12$, then $f(t)=f(\frac 12)=f(0)=i_p(1)=1!$ because $f$ is not constant.
Therefore $i_p*f\neq f$.
Could someone please check my attempts and help me with (ii) and (iii)?
Thank you.
 A: Your solution to (i) seems fine to me!
(ii) Write it explicitly as
$$
(f*g)^{-1}(t)=(f*g)(1-t)=\begin{cases}
f(2-2t)\quad\text{if $1-t\leq\frac12$}\\
g(2-2t-1)\quad\text{if $1-t\geq\frac12$}\end{cases}=\begin{cases}
g(1-2t)\quad\text{if $t\leq\frac12$}\\
f(2-2t)\quad\text{if $t\geq\frac12$}\end{cases}.
$$
(iii) Simply a circle, because moving clockwise then counterclockwise is not the same as vice versa. In fact, every closed path $f$ such that $f^{-1}\neq f$ will do. Because if $f^{-1}(t)\neq f(t)$ then $(f^{-1}*f)(\frac12t)\neq(f*f^{-1})(\frac12t)$. So say $f(t)=(\cos 2\pi t,\sin 2\pi t)$ -- this is my personal favorite example.
(iv) I think you have a good idea, but I don't get your proof. $i_p(1)=1$? My (probably similar) idea: from $f([0,\frac12])=\{p\}$ it follows that $f([\frac12,\frac34])=\{p\}$ and thus $f([\frac34,\frac78])=\{p\}$ and so on. You could say that $i_p*f=f$ means that $f$ on $[\frac12,1]$ is the same as $f$ on $I$. But this implies that whatever $f$ does, $f$ does it again twice as close to $t=1$. That's where the argument comes from. Assuming that the topological space is Hausdorff, $f([0,1))=\{p\}$ contradicts the fact that $f$ is not constant.
