Let $X$ be a topological space and $x_0 \in X.$ Given $[g], [h] \in \pi_1 (X, x_0)$ define a map $H : I \times I \longrightarrow X$ by
$$H (t,s) = \begin{cases} h \left (\frac {3t} {1 – s} \right ) & \text {if}\ 0 \leq t \leq \frac {1 – s} {3} \\ g \left ( \frac {3t + s -1} {1 + 2 s} \right ) & \text {if}\ \frac {1 – s} {3} \leq t \leq \frac {2 + s} {3} \\ \overline h \left (\frac {3t – s – 2} {1 – s} \right ) & \text {if}\ \frac {2 + s} {3} \leq t \leq 1 \end{cases}$$
Doesn't it mean that $h \ast g \ast \overline h$ is based homotopic to $g$ in $X\ $? But if it's the case then this is true for any two loops $g$ and $h$ in $X$ based at $x_0.$ But this in turn implies that $\pi_1 (X, x_0)$ is abelian, which is not necessarily true. Where did I go wrong?
Any help would be highly appreciated. Thanks for your time.
EDIT $:$ The homotopy diagram I mentioned is the following $:$
Remark $:$ Another point is that although $h \ast g \ast \overline h$ is not homotopic to $g$ as based loops in $X$ they are indeed homotopic as non-based loops in $X.$ The following gives one such homotopy. Consider $K : I \times I \longrightarrow X$ defined by
$$K (t,s) = \begin{cases} \overline h (s -3t) & \text {if}\ 0 \leq t \leq \frac {s} {3} \\ g \left ( \frac {3t – s} {3 – 2 s} \right ) & \text {if}\ \frac {s} {3} \leq t \leq \frac {3 – s} {3} \\ \overline h (3t + s – 3) & \text {if}\ \frac {3 – s} {3} \leq t \leq 1 \end{cases}$$
Best Answer
The question was answered in the comments. This community wiki summarizes the relevant insights.
Formally the homotopy $H$ is not well-defined because it involves a division by $0$ for $s=1$. It is impossible to repair the definition of $H$ as we shall explain now.
The "homotopy diagram" explains what the homotopy from $h *g * \overline h$ to $h$ should do.
The problem is that at $s=1$ the parametric interval of $h$ and $\overline h$ just becomes a point. So $h$ and $\overline h$ do not fit into this degenerate interval at $s=1$ unless $h$ is a constant loop (and consequently so is $\overline h$). For any other $0≤s<1$ we are fine because in that case the parametric intervals of $h$ and $\overline h$ are both non-degenerate closed and bounded intervals and any such interval is homeomorphic to $[0,1]$. For the degenerate case we no longer have that homeomorphism. So the insertion of $h$ and $\overline h$ into the shrinked intervals miserably fails at $s=1$.