# Hatcher's proof of the van Kampen Theorem (injectivity of $\Phi$ – unique factorizations of $[f]$)

I am trying to understand the details of Allen Hatcher's proof of the Seifert–van Kampen theorem (page 44-6 of Algebraic Topology).

My question is regarding the same part of the proof mentioned in this answer which I copy below for convenience:

In the previous paragraph, Hatcher defines two moves that can be performed on a factorization of $$[f]$$. The second move is

Regard the term $$[f_i]\in\pi_1(A_\alpha)$$ as lying in the group $$\pi_1(A_\beta)$$ rather than $$\pi_1(A_\alpha)$$ if $$f_i$$ is a loop in $$A_\alpha\cap A_\beta$$.

Regarding this move, Hatcher asserts that

[This move] does not change the image of this element in the quotient group $$Q=\ast_\alpha\, \pi_1(A_\alpha)/N$$, by the definition of $$N$$

This is the step at which Hatcher is using the hypothesis that $$N$$ is normal. In particular, if $$N$$ were simply the subgroup generated by the elements $$i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{-1}$$ (instead of the normal subgroup generated by these elements), this move would not necessarily preserve the image of this element in $$G/N$$.

I do not follow why this move does not change the image of the element in the quotient group $$Q$$. As I understand it, we have some word in $$\ast_\alpha\pi_1(A_\alpha)/N$$, (say) $$[f_1][f_2]\cdots[f_k]$$, and observe that one of the letters lies in the intersection of two of the groups in the free product, i.e. $$[f_i]\in\pi_1(A_\alpha\cap A_\beta)$$.

This means that $$i_{\alpha\beta}(f_i)i_{\beta\alpha}(f_i)^{-1}$$ is one of the generators of $$N$$. But why does it follow that changing the representative of $$[f_i]$$ in the word $$[f_1][f_2]\cdots[f_k]$$ does not affect the coset $$N[g]$$ in which $$[f_1][f_2]\cdots[f_k]$$ lies? In reply to the answer quoted above, what can go wrong if $$N$$ is not normal?

• If $N$ is not normal, then $G/N$ is not a group.
– user17892
Commented Apr 22, 2016 at 9:03
• @JustinYoung Oh yes I see now, thankyou. And the other property (the move not changing the element in the quotient group) -- I suspect that this is a property of normal subgroups (generally) as well? Commented Apr 22, 2016 at 11:20
• That is a specfic property of this $N$ (though generally we are talking colimits here), $N$ is defined precisely so that elements that come from an intersection are identified in the quotient.
– user17892
Commented Apr 22, 2016 at 11:51
• @JustinYoung Hmm I think I see that, but the elements from the intersection may fall in the middle of the word? Writing $i_{\alpha\beta}([f_i])=[f_i]_\alpha$ and $i_{\beta\alpha}([f_i])=[f_i]_\beta$; I do not understand why this equality necessarily holds: $$[f_1]\cdots[f_{i-1}][f_i]_\alpha[f_{i+1}]\cdots[f_k]=[f_1]\cdots[f_{i-1}][f_i]_\beta[f_{i+1}]\cdots[f_k].$$ The full word isn't in $N$, but $N$ is generated by words of the form $[f_i]_\alpha[f_i]_\beta^{-1}$. Commented Apr 22, 2016 at 12:33
• Ok, that's a general issue again, if you have elements $a, b$ so that $ab^{-1} \in N$, a normal subgroup in a group $G$, then show that $\overline{xay} = \overline{xby}$ in $G/N$ for any $x, y \in G$. It's not hard if you understand that $G/N$ is a group and treat it as such.
– user17892
Commented Apr 22, 2016 at 14:27

$N$ is normal if and only if $G/N$ is a group.
In any group $G$ with normal subgroup $N$, if $ab^{-1} \in N$, then $\overline{xay} = \overline{xby}$ in $G/N$ for any $x, y \in G$. This follows by cancelling the $\overline{x}$ and the $\overline{y}$ and then noting that $\overline{n} = e$ in $G/N$ for any $n \in N$, so $ab^{-1} \in N$ implies $\overline{a} = \overline{b}$ in $G/N$.