Do path homotopy classes of concatenated paths have a middle fixed point?

Question

If $[a]$ and $[b]$ are path homotopy classes, then $[a]\cdot[b]$ is defined as $[a*b]$, where $a*b$ is defined as the concatenation of the paths $a$ and $b$.

Let us say $a(1)=b(0)=p$. Then does each path in $[a*b]$ have to contain $p$?

Answer 1

No they don't have to contain $p$. That is, $[a\ast b]$ is the homotopy class of paths relative to $\{0, 1\}$, not relative to $\{0, \frac{1}{2}, 1\}$. It is very important that this is the case.

In defining the fundamental group, the inverse of $[a]$ is $[\bar{a}]$ where $\bar{a}$ is the loop $a$ traversed in reverse, i.e $\bar{a}(t) = a(1-t)$. If $[a\ast\bar{a}]$ was the set of homotopy classes relative to $\{0, \frac{1}{2}, 1\}$, then $[a\ast\bar{a}]$ would not in general be the homotopy class of the constant loop; that is, $a\ast\bar{a}$ is not in general homotopic to the constant loop relative to $\{0, \frac{1}{2}, 1\}$ (see the gif below from Wolfram MathWorld to see why).

$\hspace{68mm}$

As the homotopy class of the constant loop is the identity of the fundamental group, we see that $[\bar{a}]$ would not be the inverse of $[a]$.