In Munkres, page 326 on Fundamental Groups, we have $$ [f] \star [g] = [f \star g]$$ where $f$ is a path from $x_0$ to $x_1$, and $g$ is a path from $x_1$ to $x_2$. $\star$ defines the composition of the two paths. $[f]$ is the path-homotopy equivalence class of $f$. Now, I am just learning this concept, so please let met know if I make a mistake in my wording below!
I am unable to see why $[f \star g] \subset [f] \star [g] $. For example, consider a sample $f_0 \star g_0$ - it is a path from $x_0$ to $x_2$ that goes through $x_1$. Now, this could be homotopically equivalent to another path $h_1$ from $x_0$ to $x_2$ that doesn't goes through $x_1$.
$h_1 \in [f \star g]$, therefore.
However, by the above equality, should there exist $f_1, g_1 \in [f], [g]$, respectively, such that $h = f_1 \star g_1$? If so, shouldn't this be impossible, since every such composition must cross through $x_1$? Or do they not have to be identically equal, but rather, that there exists a homotopy between $h$ and $f_1 \star g_1$?
In Munkres' proof of the inequality, he shows that $\forall f, f' \in [f]$ and $\forall g, g' \in [g]$, such that F is the homotopy between $f, f'$ and G is the homotopy between $g, g'$, $F \star G$ is the homotopy between $f \star g$ and $f' \star g'$. That is, $f' \star g' \in [f \star g]$, so that $ [f] \star [g] \subset [f \star g]$. Why is this a proof of the above equality? Where is my understanding wrong?