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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?

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Szmagpie
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    If $N$ is not normal, then $G/N$ is not a group. – Justin Young Apr 22 '16 at 09:03
  • @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? – Szmagpie Apr 22 '16 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. – Justin Young Apr 22 '16 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 $[f_i]\alpha[f_i]\beta^{-1}$. – Szmagpie Apr 22 '16 at 12:33
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    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. – Justin Young Apr 22 '16 at 14:27
  • @JustinYoung Aha I see! Because $\overline{a}={na:n\in N}={(mba^{-1})a:m\in N}=\overline{b}$ since for all $n\in N$ there is unique $m\in N$ with $mba^{-1}=n$ (choosing $m=nab^{-1}$). $$\overline{a}=\overline{b}\Rightarrow\overline{xay}=\overline{xby}.$$ Appreciate the help, if you would like to write an answer, I will accept, else I will post it myself. :-) – Szmagpie Apr 22 '16 at 17:18

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$N$ 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 $G/N$.

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