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This comes from the application of Van-Kampen theorem. Note Van-kampen theorem, states for $U$, $V$ and $U \cap V$, are open and path connected space we have \begin{align} \pi_1 (U \cup V) =\pi_1 (U) *_{\pi_1(U\cap V)} \pi_1(V) \end{align} Here what i want to prove is for $U\cap V$ is simply connected, \begin{align} \pi_1 (U\cup V) = \pi_1(U)* \pi_1(V) \end{align}

First, starting from the definition of amalgamted free product \begin{align} \pi_1 (U) *_{\pi_1(U\cap V)} \pi_1(V) = \pi_1(U) * \pi_1(V) / K \end{align} where $K$ is normal sub-group generated by $ \{ f_1(a) f_2(a)^{-1}|a\in \pi_1(U\cup V)\}$ where $f_i$ are homomorphism, $f_i: N \rightarrow G_i$ for $i=1,2$

The professor says, $K$ is trivial since $\pi_1(U\cap V)$ is trivial. Which i did not fully understand yet.

I know that $\pi_1(U\cap V)$ is trivial since $U\cap V$ is simply connected. Recall the meaning of trivial, it has a element, identity. Thus i can see $a$, which is a element of $\pi_1(U\cap V)$ is identity.

But i am not sure why $K$ is trivial. Can you give me some detail explanation for this?

B. Pasternak
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phy_math
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    It's actually written in your question; since $\pi_1(U\cap V)$ is trivial, we have ${f_1(a)f_2^{-1}(a),|,a\in\pi_1(U\cap V)}={f_1(e)f_2^{-1}(e)}={e}$ since $f_i$ is an homomorphism, so $K$ is actually trivial. – B. Pasternak May 13 '16 at 14:39
  • @B.Pasternak, can you explain more, about the process, between the $f_i$ is an homomorphism and $K$ is trivial? – phy_math May 13 '16 at 14:44
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    Since $f_i$ is an homomorphism for $i=1,2$, $f_i(e)=e$ (in your question you write $f_2^{-1}(a)$; this is wrong, it should be $(f_2(a))^{-1}$). – B. Pasternak May 13 '16 at 14:47
  • @B.Pastemak. Oh i see, thanks!! – phy_math May 13 '16 at 14:50

1 Answers1

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$K$ is the normal subgroup generated b $\{f_1(a)(f_2(a))^{-1}, a\in\pi_1(U\cap V)\}$, thus is trivial since $\pi_1(U\cap V)$ is trivial.

B. Pasternak
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Tsemo Aristide
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