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Let $X$ be the space obtained from the torus $S^1 \times S^1$ by attaching a Mobius band via a homeomorphism from the boundary circle of the Mobius band to the circle $S^1 \times \{ x_0 \}$ in the torus. Compute $\pi_1(X)$.

We use Van Kampen theorem, letting $M$ and $T$ denote the Mobius band and the torus respectively. Note that $M \cap T \simeq S^1$ and therefore, we have an isomorphism $$\frac{\pi_1(M,x_0) \ast \pi_1(T,x_0)}{N} \cong \pi_1(X,x_0),$$ where $N$ is the normal subgroup generated by $i_{MT}i_{TM}(\omega)^{-1}$ for $\omega \in \pi_1(M \cap T, x_0)$.

I'm unsure of what this $N$ is exactly.

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See, in Van Kampen's theorem, we are finding the fundamental group of a topological space, given that it is the union of spaces, whose fundamental group is known to us.

This is done as follows: Imagine two sets $U$ and $V$, whose fundamental group you know, and imagine that they follow the conditions for Van Kampen theorem.

What essentially happens in Van Kampen theorem? Well, imagine you want to create a loop in $U \cup V$. How would you do it? Start from the basepoint, you can go through a loop either in $U$ or $V$ or both. Come back, again you can go to a loop either in $U$ or $V$ or both. You can repeat loops, you can go the wrong way around a loop, and this way, a loop in $X$ is created.

For example, if $U$ has a loop $a$ and $V$ has a loop $b$, then a typical loop in $X$ would be read off as $a^2ba^{-1}b^{-2}ab^4$, which means "go round $a$ twice, go round $b$ once, go the wrong way round $a$ once, $\ldots$ go round $b$ four times".

These elements, created using the smaller loops $a$ and $b$, is called the free group generated by $a$ and $b$. According to Van Kampen, the fundamental group of $X$ is this free group,quotiented by something. But what?

Now, what is happening with the loops in the intersection $U \cap V$? To understand this, see that loops in the intersection $U \cap V$ belong to fundamental groups of both $U$ and $V$.

Now, what happens during the construction of the free group I mentioned above, is that (I'll try to make this as intuitionistic as possible, because the free group is not an easy concept) this one loop ends up getting counted twice, once as an element of $U$ and as an element of $V$. To give a very weak analogue, it is like this: $2$ is one element, but $(0,2)$ and $(2,0)$ are different elements, like that.

The quotient precisely prevents this from happening. To see how this is happening, $N$ is doing the following: $i_{MT}$ is the inclusion map of loops in $M \cap T$ into loops in $M$, or just the inclusion $\pi_1(M \cap T, x_0) \to \pi_1 (M,x_0)$, and similarly, $i_{TM}$ is the inclusion map of loops in $M \cap T$ into loops in $T$, or just the inclusion $\pi_1(M \cap T, x_0) \to \pi_1 (T,x_0)$. Basically, the loop remains unchanged, but the inclusion map changes the fundamental group it is part of.

So what do we do to prevent this? We multiply this loop by it's own inverse, by doing the operation $(i_{MT}(w)i_{TM}(w)^{-1})$, changing the loop to an identity loop, which can be discounted. Now, these elements ( think of $(0,2)$ and $(2,0)$) become related by the equivalence relation generated by the loop's multiplication with it's own inverse, and hence all of them fall into one equivalence class (which we can treat as $[(0,2)] = [(2,0)]$, now they are not distinct!). Hence, the same loop belonging to two or more fundamental groups at once, gets counted precisely once!

And the beauty is, this is a proof, almost without symbols, of Van Kampen's theorem, because every loop in $X$ can be written using the free group logic I described earlier (surjective) and because of quotienting, every such common loop is counted precisely once (injectivity).

Thus, the quotienting is required to preserve the isomorphism, and is very integral to Van Kampen's theorem.

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  • $\begingroup$ By the acceptance, I take it you have understood the question. Have you been able to calculate $N$ is the question above now? Van Kampen's theorem can be applied to the question, it has a nice answer. $\endgroup$ Nov 5 '16 at 1:18
  • $\begingroup$ @ астон вілла олоф мэллбэрr We let $\omega \in \pi_1(M \cap T, x_0)$ and consider that $\pi_1(M \cap T, x_0) \cong \mathbb{Z}$, so the loops that end up being redundant are those that wrap around this $S^1$. With $\mathbb{Z}$ having only 1 generator, we have that $N \cong \mathbb{Z}$. So the fundamental group is given by $\mathbb{Z} \ast (\mathbb{Z} \times \mathbb{Z})/\mathbb{Z} \cong \mathbb{Z} \ast \mathbb{Z}$? $\endgroup$
    – user319128
    Nov 5 '16 at 1:24
  • $\begingroup$ @Elliot Unfortunately I'm not very exposed to isomorphisms of free groups etc. , but surely $\mathbb Z * (\mathbb Z \times \mathbb Z) / \mathbb Z$ is correct, unfortunately I can't say confidently that it is isomorphic to $\mathbb Z *\mathbb Z$. If you can ,then you are correct. $\endgroup$ Nov 5 '16 at 1:27
  • $\begingroup$ @ астон вілла олоф мэллбэрr Thank you every much! $\endgroup$
    – user319128
    Nov 5 '16 at 1:31
  • $\begingroup$ @Elliot You are welcome. $\endgroup$ Nov 5 '16 at 1:34

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