I am trying to understand the proof to the Van Kampen theorem and an important step in it is decomposing a topological space $X$ into the pushout of its open subspaces $A,B$. It is relatively easy to see that $X$ with the maps $f_A:A\to X,f_B:B\to X$ has the universal property of the pushout and hence is homeomorphic to $A+_{A\cap B}B$ (as I understood it, this basically comes down to the gluing lemma and the uniqueness of the pushout), but it is stated in the proof that the condition that $A,B$ be open in $X$ is strictly necessary and it is suggested that one try and come up with an example as to why. I am having trouble coming up with one, is there a standard example that illustrates why this condition is necessary?
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1$\begingroup$ For example, the pushout of $(-\infty, 0]$ and $(0, \infty)$ with subspace topologies from $\mathbb{R}$ is not homeomorphic to $\mathbb{R}$, but rather it's homeomorphic to $(-\infty, 0] \cup (1, \infty)$. $\endgroup$Daniel Schepler– Daniel Schepler2021-09-30 16:45:29 +00:00Commented Sep 30, 2021 at 16:45
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$\begingroup$ @DanielSchepler I tried following your first comment and considered the pushout of $[0,\frac{1}{2})$ and $[\frac{1}{2},1]$, which can't be homeomorphic to $[0,1]$, since otherwise there'd be a non-constant map from it to $\{0,1\}$. This seems to work, but it still feels like a "stilted" example (since the intersection is empty) and I'm trying to understand this more conceptually. Can you offer a more conceptual explanation as to why the sets should be open? $\endgroup$St. Barth– St. Barth2021-09-30 16:49:11 +00:00Commented Sep 30, 2021 at 16:49
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2$\begingroup$ All examples will feel kind of weird because there are plenty of instances where the subspaces aren't open but van Kampen still works. The point of the premises are to ensure weird things don't happen, so any counter examples are going to be weird. $\endgroup$Connor Malin– Connor Malin2021-09-30 16:52:58 +00:00Commented Sep 30, 2021 at 16:52
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$\begingroup$ Another counterexample where the two subspaces do intersect nontrivially: in the circle $S^1 \simeq \mathbb{R} / 2\pi \mathbb{Z}$, consider the subspaces $[0, 3\pi/2)$ and $(\pi/2, 2\pi)$. The pushout of the inclusion maps from the intersection $(\pi/2, 3\pi/2)$ is then homeomorphic to the interval $[0, 2\pi)$ of $\mathbb{R}$, not to $S^1$. $\endgroup$Daniel Schepler– Daniel Schepler2021-09-30 16:55:12 +00:00Commented Sep 30, 2021 at 16:55
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"It is stated in the proof (of the Seifert - van Kampen theorem) that the condition that $A,B$ be open in $X$ is strictly necessary."
This is an ill-chosen expression. That $A,B$ be open in $X$ is a sufficient condition. There are many examples where the theorem is true altough $A,B$ are not open. So "$A,B$ open" is not a necessary assumption the logical sense.
What is meant is that without assuming $A,B$ open the given proof breaks down. And in fact, the Seifert - van Kampen theorem is not true without certain assumptions on $A,B$.
The theorem in its standard form says that in all counterexamples at least one of $A,B$ must be non-open. Here is such a counterexample:
In the unit circle $S^1 \subset \mathbb R^2$ let $A$ denote the closed upper half circle, $B$ the open lower half circle and $C$ the open right half circle. Define $U_1 = A \cup C$ (which is not open) and $U_2 = B \cup C$ (which is open). We have $U_1 \cup U_2 = S^1$, $U_1 \cap U_2 = C$. The intersection is open and path-connected and with $p = (1,0) \in C$ we get $$\pi_1(U_1,p) = \pi_1(U_2,p) = \pi_1(U_1 \cap U_2,p) = 0 .$$ The Seifert - van Kampen theorem fails - otherwise we would have $\pi_1(S^1,p) = 0$.
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$\begingroup$ it was poorly worded on my part, it's not like this exactly in the proof. The "necessity" of the openness condition comes up in an exercise that asks one to prove that a space can be realised as the pushout of its open subsets. My question had more to do with why exactly it is necessary at this junction to assume this, but your example illustrates it nicely from the point of view of the theorem itself. Thanks. $\endgroup$St. Barth– St. Barth2021-10-01 19:47:19 +00:00Commented Oct 1, 2021 at 19:47