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I am studying Van Kampen Theorem using Hatcher's textbook. I am dealing with the general statement, I mean: (pg 43)

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He defines previously the free product of groups (pg 41) as:

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I can follow the main idea of the proof but I don't understand how he can say (pg 45):

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By definition, elements of the free product should be reduced words, am I right? Then he should not be considering unreduced words in a free product, since there won't be none. I suppose I have missed something, if that is the case and I am pretty sure that it is, Where I am misunderstanding?

Thanks in advance!

D1811994
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1 Answers1

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The author might need to consider unreduced words in order to simplify the presentation of the proof. So while, technically speaking, it is slight abuse of language to say that these unreduced words live in the free product, one may extend the map $\Phi$ to these words by reducing them first and then applying the original $\Phi$.

Alex Provost
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  • The question is near to a duplicate of http://math.stackexchange.com/questions/136518/a-question-about-the-proof-of-seifert-van-kampen?rq=1 . There is also a case to be made for proving the Seifert-van Kampen Theorem by verifying the universal property, as in the proof by 1958 Crowell . – Ronnie Brown Feb 28 '16 at 11:11
  • @RonnieBrown thanks for pointing out other way of proving the theorem. It could seem near to a duplicate but the user who asked that question assumed what I am asking. I mean, I had read that question and answers before asking mine and they hadn't clarified my doubt. Thanks anyway for referencing to that question! By the way, in the end I suppose I would end by reading your proof: http://pages.bangor.ac.uk/~mas010/pdffiles/brown-razak.pdf – D1811994 Feb 28 '16 at 13:06
  • The nice thing about the proof of VKT by universal property is that it proves this particular colimit exists, but does not assume general colimits exist. This is useful when moving for example to higher dimensions. Nor does this method assume any particular way of constructing colimits.It may be that your question is more about the latter, for the case of coequalisers. – Ronnie Brown Feb 28 '16 at 18:09