show "product operation on paths induces a well defined operation on path homotopy classes" This is from Munkres, Topology. I want to show that for $f,g \in \Omega(X,x_0)$, which is the set of all loops at base point $x_0$, that the following holds:
$$[f]*[g]=[f*g]$$
where $*$ is the usual binary operation defined on paths. 
My source of confusion stems from interpreting the actual meaning of $=$ in the previous relationship. It is one thing to say that $[f]=[g]$ if $f\simeq g$ via some homotopy. On the other hand, I am under the impression that $[f]*[g]$ is also a set consisting of the product of elements in the two equivalence classes i.e. I am under the impression that $[f]*[g]:=\{j*k: j\in [f], k\in [k]\}$ from this it would be necessary to show that $h*m \in [f]*[g], h*m \in [f*g] $ and vice versa. 
However the proof that Munkres provides only deals with showing that the product of path homotopies on the left side is the required path homotopy for the product of paths on the right hand side.
What is the formal interpretation of $[f]*[g]$ and how does it relate to the proof provided by Munkres?
 A: Munkres defines $[f]*[g] = [f * g]$ and proves that this is well-defined. In fact, he shows
$$\{ f' * g' \mid f' \in [f], g' \in [g] \} \subset  [f* g] .$$
In general you do not have $\{ f' * g' \mid f' \in [f], g' \in [g] \} = [f]*[g]$. As an example take $(X,x_0) = (S^1,1)$, where $S^1 = \{ z \in \mathbb{C} \mid \lvert z  \rvert = 1 \}$. Let $f(t) = e^{2\pi i t}$ and $g(t) = f(1-t) = e^{-2\pi i t}$. Then $[f] * [g] = [e]$, where $e$ is the constant path $e(t) = 1$. However, there do not exist $f' \in [f], g' \in [g]$ such that $f' * g' = e$. This would only be possible if $f' = g' = e$, but it is well-known that $[f] \ne [e]$ and $[g] \ne [e]$.
The equation $[f]*[g] = \{ [f' * g'] \mid f' \in [f], g' \in [g] \}$ in your comment is also wrong. On the left hand side you have a set of paths, on the right hand side a set of sets of paths. 
A: What you need to show for well-definedness of $[f] \ast [g]$ is that if we take any representative $f' \simeq f$ and any $g' \simeq g$, then $f' \ast g' \simeq f \ast g$, i.e. we get the same homotopy class regardless of representatives. This common class is then defined to be the product.
The proof is on page 2 of my note here, where I also check all the axioms.
