# A question about the proof of Seifert - van Kampen

I'm studying Algebraic Topology using Hatcher's textbook as my main reference, and there is a detail in the proof of the Seifert - van Kampen Theorem (page 43) which is still unclear to me. The heart of the proof of the second part of the statement (the one regarding relations) is where it says (page 45):

If we can show that any two factorizations of $[f]$ are equivalent, this will say that the map $Q \rightarrow \pi_1(X)$ induced by $\phi$ is injective, hence the kernel of $\phi$ is exactly $N$, and the proof will be complete.

I'm not exactly sure which result Hatcher is referring to here, but the way I see it, this statement is a sort of a converse of the classical homomorphism theorem (if $\phi: G \rightarrow G'$ is a group homomorphism then $\phi$ canonically defines an injective map ${G}/{\ker \phi} \rightarrow G'$). To make things rigorous I stated the following

Lemma Let $\phi: G \rightarrow G'$ be a group homomorphism and $N \leq \ker \phi$ ($N$ not necessarily normal in $G$). Let $G/N^r$ be the set of right cosets of $N$. Then we can define \begin{align} \tilde{\phi}: G/N^r &\rightarrow G' \\ Nx &\mapsto \phi(x) \end{align} and if $\tilde{\phi}$ is injective then $\ker \phi = N$.

I managed to prove this lemma (if necessary I will edit the question to include such proof), but the problem now is that this seems to be too strong a result! What the Theorem says is that the kernel of $\Phi$ is the normal subgroup generated by all the $i_{\alpha \beta}(\omega) i_{\beta \alpha}(\omega)^{-1}$'s; it doesn't say that the kernel is the subgroup generated by all such elements, which also happens to be normal. By applying the above lemma to Hatcher's proof, with $G = \ast_{\alpha} \pi_1({A_\alpha)}, N = \langle \{ i_{\alpha \beta}(\omega) i_{\beta \alpha }(\omega)^{-1}: \omega \in \pi_1 (A_{\alpha} \cap A_{\beta}) \}\rangle, \phi = \Phi$ (note that $N \leq \ker \phi$ is remarked at the beginning of page 43), and by subsituting every instance of $Q = \ast_{\alpha} \pi_1(A_{\alpha})$ with the set of right cosets, one would thus be able to prove a stronger version of the theorem (taking $\ker \Phi$ to be the subgroup generated by the $i_{\alpha \beta}(\omega) i_{\beta \alpha }(\omega)^{-1}$'s , and not its normal closure; the normality of such a subgroup would then be a consequence of the fact that it is the kernel of a group homomorphism). This makes me suspicious that I might be getting something wrong...
So, to summarize, here are my questions:

• What am I getting wrong? Have I misunderstood the statement of the theorem, is the lemma wrong or is the normality of the subgroup used in some other part of the proof? Or is everything right?

• Can someone provide an explicit example in which the "stronger version" of the theorem doesn't hold: i.e. in which $\langle \{ i_{\alpha \beta}(\omega) i_{\beta \alpha }(\omega)^{-1}: \omega \in \pi_1 (A_{\alpha} \cap A_{\beta}) \}\rangle$ is different from $\langle \{ i_{\alpha \beta}(\omega) i_{\beta \alpha }(\omega)^{-1}: \omega \in \pi_1 (A_{\alpha} \cap A_{\beta}) \}^{\ast_{\alpha} \pi_1(A_{\alpha})} \rangle$.

Thank you very much for any help.

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The right cosets are often denoted by $N\backslash G$, by the way. –  Dylan Moreland Apr 25 '12 at 13:46
Thanks, I looked into the standard notations for the set of right/left cosets, and changed it to $G/N^r$. –  Emilio Ferrucci Apr 25 '12 at 14:11
Oddly enough, I've never seen that notation. –  Dylan Moreland Apr 25 '12 at 14:14
Neither had I, I found it here proofwiki.org/wiki/Definition:Coset_Space , there doesn't really seem to be a completely unambiguous notation.. –  Emilio Ferrucci Apr 25 '12 at 14:21

## 3 Answers

I think the flaw in your reasoning comes earlier in the proof. 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 $G/N$.

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Thank you for your answer. I still don't understand one thing though: why does simply substituting $Q$ with the set of left cosets not work? No group structure on $Q$ is used to complete the proof, and a group can still be partitioned in cosets modulo a nononrmal subgroup. Doesn't the second remark you quoted just use the fact that two such factorizations represent the same coset? –  Emilio Ferrucci Apr 25 '12 at 10:00
It is not true in general that two such factorizations represent the same coset. If $N\leq G$ is not normal and $aN=bN$ and $cN=dN$, it does not follow that $acN = bdN$ –  Jim Belk Apr 25 '12 at 14:25
Ah, I see.. the well-definedness of the coset multiplication is used after all. My reasoning only held for words of length $1$. Thanks a lot for clearing this up for me. –  Emilio Ferrucci Apr 25 '12 at 14:47

I think it should be pointed out that another style of proof is available, and is used to give a proof of the version using many base points, and so groupoids, at

http://www.bangor.ac.uk/r.brown/pdffiles/vKT-proof.pdf

With the result for many base points, you can compute the fundamental group of the circle, which is, after all, THE basic example in algebraic topology. The proof given there does only the union of 2 open sets, but it gives the proof by

verification of the universal property,

which is a general procedure of great use in mathematics. For example this method is used to prove higher dimensional versions of the van Kampen Theorem. This method also avoids description of the result by generators and relations.

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Thank you for your interest. I can't open the website you linked though.. –  Emilio Ferrucci Apr 25 '12 at 18:20
Thanks for pointing this out - my mistake, now corrected! –  Ronnie Brown Apr 29 '12 at 20:11

Ok, I can at least show why if $N$ is a normal subgroup of $*_{\alpha}\pi_{1}(A_{\alpha})$, $N\subseteq \ker\Phi \subseteq *_{\alpha}(A_{\alpha})$, and $\Phi:*_{\alpha}\pi_{1}(A_{\alpha})\rightarrow\pi_{1}(X)$ is injective, then $\ker\Phi=N$.

The reason is as follows: we have the following diagram


and the mapping $f$ that makes the diagram commutes is the following:

$$\begin{eqnarray} f:&*_{\alpha}\pi_{1}(A_{\alpha})/N\rightarrow *_{\alpha}\pi_{1}(A_{\alpha})/\ker\Phi\\ & gN\rightarrow g\ker\Phi \end{eqnarray}$$

By the fact that $\Phi:*_{\alpha}\pi_{1}(A_{\alpha})/N\rightarrow \pi_{1}(X)$ is injective, this makes $f$ an injective homomorphism. On the other hand, $f$ is also surjective. By a theorem in group theory (3rd isomorphism theorem), $\ker f=(\ker\Phi)/N$. But $f$ is also injective, so $(\ker\Phi)/N=\{N\}$. Hence if $x\in\ker\Phi$, then $xN=N$ means that $x\in N$. So $\ker\Phi=N$.

I don't understand is why

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 $A_{\alpha}\cap A_{\beta}$.

would imply that

[This move] does not change the image of the element in $Q$...

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