# geometric argument for van-kampen?

I've seen Van-Kampen's theorem presented algebraically many times; and although it provides a useful method of calculation; I don't have a very clear picture for "why" it should be true. Does anyone know of a more visual argument; or even an example that makes the it easier to see?

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Can you visualize the case where the intersection is simply connected? – Qiaochu Yuan Feb 7 '11 at 10:48
Have you read the proof? The surjectivity part gives you the geometrical insight. – Gabriel Furstenheim Feb 7 '11 at 15:15
I think the proof in Hatcher is quite geometric/visual. – Soarer Feb 7 '11 at 15:55

## 2 Answers

I'm only going to try to give some intuition why the Seifert-van Kampen Theorem is geometrically reasonable, rather than attempt a geometric proof (which would be well beyond my abilities, I fear).

The Seifert-van Kampen Theorem ("van Kampen" was the last name, so it should not be hyphenated) says that if $X=U\cup V$, with $U$ and $V$ arcwise connected open subsets, and $U\cap V$ is nonempty and arcwise connected, then choosing a basepoint $x_0\in U\cap V$, then $$\pi(X) \cong \pi(U)*_{\pi(U\cap V)}\pi(V);$$ that is, the fundamental group of $X$ is the free amalgamated product of the fundamental groups of $U$ and $V$, amalgamated over the subgroup corresponding to the fundamental group of $U\cap V$.

Intuitively: it should be clear that $\pi(X)$ contains subgroups isomorphic to $\pi(U)$ and $\pi(V)$, and since $X=U\cup V$, with both open, it should be reasonable that $\pi(X)$ is generated by these two subgroups: if you have an arbitrary loop in $X$ that meanders between $U$ and $V$, you can try decomposing it into an equivalent loop that alternates being entirely in $U$ and entirely in $V$. Also, $\pi(V)$ and $\pi(U)$ will each have subgroups corresponding to $\pi(U\cap V)$. So the real question is why this should be the only interaction between $\pi(U)$ and $\pi(V)$ (that is, why you get the free product, and not just some quotient of it).

A relation would correspond to some loop in $X$ which goes through $U$, then $V$, then $U$, then $V$, etc., and is homotopy equivalent to the trivial loop. Imagine that you find such a loop, and pick one with the smallest possible number of "crossings" from one set to the other which satisfies this. You can imagine setting up your deformation so that it first "shrinks" the "last" part, until it is completely contained in $U\cap V$ (since this is open and that is where the basepoint is). But that would give you a new loop which is also equivalent to the identity, but with a fewer number of "crossings" (since you can call the portion in $U\cap V$ part of $U$ or part of $V$, as convenient). This tells you that the smallest possible number of crossings has to be $0$; that is, the entire thing was contained inside $U\cap V$ in the first place. So you should not expect any relations between the $\pi(U)$ and $\pi(V)$, except those that come from identifying their common subgroup $\pi(U\cap V)$. This is precisely the free product with amalgamation.

A good place to try to get some intuition is the case where the intersection is simply connected, like Qiaochu suggests in the comments; I find bouquets of circles particularly fertile ground, but then I'm a group theorist, so this is almost the only kind of fundamental group I play with on a semi-regular basis.

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I will say it less acurately than Arturo and sure enough too much roughly. Consider it as a first approxiation to the theorem.

The way I see Seifert-Van Kampen theorem is like a sophisticated version of the dimension of the sum of vector subspaces:

$$\mathrm{dim} (F + G) = \mathrm{dim} F + \mathrm{dim} G - \mathrm{dim} (F\cap G) \ .$$

Because, forgetting torsion, the fundamental group counts the number of "holes" (non contractible loops, generators of $\pi_1$) in your space $X$. Right?

So, if you have $X = U \cup V$, then the number of holes in $X$ must be equal to the number of holes in $U$ plus the number of holes in $V$..., minus the number of holes in $U\cap V$, because these ones you have already counted them twice.

Hence, in order to count all the holes in $U$ plus those in $V$, you take the free product $\pi_1 (U) * \pi_1(V)$, but you must "identify" the holes "shared" by $U$ and $V$. This is the reason for the amalgamated product $\pi_1(U) *_{\pi_1(U\cap V)} \pi_1(V)$.

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