# Question about Hatcher's proof of van Kampen's theorem

I am reading Hatcher's proof of van Kampen's theorem. Hatcher defines what he means to be a factorization of an element $[f]$ in $\pi_1(X)$ and then defines what it means for two factorizations to be equivalent. Hatcher says that two factorizations of $[f]$ are equivalent if they are related by a sequence of moves or their inverses. For the precise definitions, I refer you to page 44 of his free online textbook Algebraic Topology.

Now, he makes the following claim (abbreviated):

If we can show that any two factorizations of $[f]$ are equivalent, this will say that the map $Q\to \pi_1(X)$ induced by $\Phi$ is injective.

Can someone prove this claim for me?

There is nothing fancy going on, or much to prove really. By the way Hatcher has defined things, if two factorizations $[f_1]\cdots[f_k]$ and $[g_1]\cdots[g_l]$ are equivalent then they determine the same element of $Q$ (which he mentions).
Suppose it is true that any two factorizations of an element $[f]$ are equivalent. To show injectivity of $\Phi$, suppose $\Phi([f_1]\cdots[f_k])=\Phi([g_1]\cdots[g_l])$, call the image $[f]\in\pi_1(X)$. By definition of $\Phi$, the elements $[f_1]\cdots[f_k]$ and $[g_1]\cdots[g_l]$ are both factorizations of $[f]$, so by our assumption they are equivalent, hence are the same as elements of $Q$. That is, $[f_1]\cdots[f_k]=[g_1]\cdots[g_l]$ in $Q$, showing injectivity of the map $Q\to\pi_1(X)$.
• Are we making implicit use of either of his two moves in this proof, or is it really as simple as writing down what $\Phi$ does to elements of $Q$? Also, perhaps the arguments of $\Phi$ in your proof should be notated to reflect that they are cosets of $N$? Commented May 8, 2017 at 21:25
• All that matters about the two "moves" is that if two elements differ by any combination of these two moves, then they determine the same element of $Q$. Nothing else about the moves themselves is being used here. And yes, maybe that would be more "correct" to use coset notation, but that'd be more difficult to read. It's standard abuse of notation to just write "equal as elements of $Q$", etc. Commented May 8, 2017 at 21:42