1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ If $N$ is not normal, then $G/N$ is not a group. $\endgroup$ – Justin Young Apr 22 '16 at 9:03
  • $\begingroup$ @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? $\endgroup$ – Szmagpie Apr 22 '16 at 11:20
  • $\begingroup$ 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. $\endgroup$ – Justin Young Apr 22 '16 at 11:51
  • $\begingroup$ @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 $[f_i]_\alpha[f_i]_\beta^{-1}$. $\endgroup$ – Szmagpie Apr 22 '16 at 12:33
  • 1
    $\begingroup$ 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. $\endgroup$ – Justin Young Apr 22 '16 at 14:27
2
$\begingroup$

$N$ 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 $G/N$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.