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$X$ is Hausdorff and locally arcwise connected, $X=A\cup B$, where $A$ and $B$ are closed subspaces of $X$, $A\cap B =\{ p\}$ and $p$ has open contractible neighborhoods $U,V$ in $A,B$ respectively. Show $\pi_1(X,p) \cong \pi_1(A,p)*\pi_1(B,p)$.

I am at a loss as to how to solve this.

Seifert-van Kampen theorem sounded okay but $A,B$ are closed and not open, so I can't use the theorem. $U,V$ are open and their intersection is simply connected but $\pi_1(U,p)$ and $\pi_1(V,p)$ are each the trivial group as they are contractible spaces. I was hoping to somehow apply Seifert-van Kampen and somehow extend this so that it covers for $X$.

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    Hint: take two singletons ${a}\subseteq A$ and ${b}\subseteq B$ which are closed subspaces and whose complements are open. That ${p}$ has a contractible neighborhood is important. –  Jan 31 '14 at 03:59

1 Answers1

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In these sorts of questions, the basic idea is to massage the situation so that your standard result (here Seifert-van Kampen) applies.

Have you tried drawing a picture of a space $X$ which is a union of two closed sets meeting in one point? If you do, perhaps you will see how to make a minor modification to $A$ and $B$ so that they each become open.

Matt E
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  • My general thoughts are we want to pick a point $a\neq p \neq b$ with $a\in A$ and $b \in B$. Then $A-a$ and $B-b$ will be open and $X'=X-{a,b}=(A-a)\cup(B-b)=A'\cup B'$ and in this case I would use the theorem to get $\pi_1(X',p)\cong \pi_1(A',p)*\pi_1(B',p)$ and I would need to show $\pi_1(X,p)\cong \pi_1(X',p)$ and similarly that $\pi_1(A,p)\cong \pi_1(A',p)$, is this a good direction to go in? Thanks for your help –  Jan 31 '14 at 04:21
  • I would think of $A'\cap B'$ more. Is it contractible? –  Jan 31 '14 at 05:14
  • @Wishingwell: Dear Wishingwell, draw two circles meeting at a point (i.e. a figure 8); these are $A$, $B$, and $p$. Now, if you carry out your suggestion, does it look like it will work? (I get the feeling that you haven't tried to draw any pictures, and are just manipulating the symbols. Try drawing the picture I suggest, and then ask if it's reasonable to expect that $\pi_1(X)$ and $\pi_1(X')$ are at all similar.) Regards, – Matt E Jan 31 '14 at 19:55
  • @Wishingwell: P.S. As a technical remark, $A\setminus {a}$ and $B\setminus {b}$ won't typically be open in $X'$, which you'll see if you draw the example I suggested. Also, if you delete a point from a space (e.g. $a$ from $A$) this is typically not a "minor modification". Think about deleting a point from a cirlce. – Matt E Jan 31 '14 at 19:57
  • Consider $A' = A\cup V$ and $B' = B\cup U$. Then, by assumption $A' \simeq A$, $B' \simeq B$, and $A' \cap B' = U\cup V \simeq p$. Also, for example, $X \setminus A' = B\setminus V$ which is closed, so $A'$ is open. Now, apply ordinary Van Kampen. – Justin Young Jun 07 '18 at 14:13