Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have just learned about the Seifert-Van Kampen theorem and I find it hard to get my head around. The version of this theorem that I know is the following (given in Hatcher):

If $X$ is the union of path - connected open sets $A_\alpha$ each containing the basepoint $x_0 \in X$ and if each intersection $A_\alpha \cap A_\beta$ is path connected, then the homomorphism $$\Phi:\ast_\alpha \pi_1(A_\alpha) \to \pi_1(X)$$ is surjective. If in addition each intersection triple intersection $A_\alpha \cap A_\beta \cap A_\gamma$ is path-connected, then $\ker \Phi$ is the normal subgroup $N$ generated by all elements of the form $i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega^{-1})$, and so $\Phi$ induces an isomorphism $$\pi_1(X) \cong \ast_\alpha \pi_1(A_\alpha)/N.$$

$i_{\alpha\beta}$ is the homomorphism $\pi_1(A_\alpha \cap A_\beta) \to \pi_1(A_\alpha)$ induced from the inclusion $A_\alpha \cap A_\beta \hookrightarrow A_\alpha$ and $\omega$ is an element of $\pi_1(A_\alpha \cap A_\beta)$.

Now I tried to get my head round this theorem by trying to understand the example in Hatcher on the computation of the fundamental group of a wedge sum. Suppose for the moment we look at $X = X_1 \vee X_2$. I cannot just apply the theorem blindly because $X_i$ is not open in $X$. So we need to look at

$$A_1 = X_1 \vee W_2, \hspace{3mm} A_2 = X_2 \vee W_1$$

where $W_i$ is a neighbourhood about the basepoint $x_1$ of $X_1$ that deformation retracts onto $\{x_1\}$, similarly for $W_2$. I believe each of these is open in $X_1 \vee X_2$ because each $A_i$ is the union of equivalence classes that is open in $X_1 \sqcup X_2$. Now how do I see rigorously that $A_1 \cap A_2$ deformation retracts onto the point $p_0$ (that I got from identifying $x_1 \sim x_2$) in $X$? If I can see that, then I know by Proposition 1.17 (Hatcher) that

$$\pi_1(A_1 \cap A_2) \cong \pi_1(p_0) \cong 0$$

from which it follows that $N= 0$ and the Seifert-Van Kampen Theorem tells me that

$$\pi_1(X_1\vee X_2) \cong \pi_1(X_1) \ast \pi_1(X_2).$$

1) Is my understanding of this correct?

2) What other useful exercises/ examples/applications are there to illustrate the power of the Seifert-Van Kampen Theorem? I have also seen that you can use it to prove that $\pi_1(S^n) = 0$ for $n \geq 2$.

I have had a look at the examples section of Hatcher after the proof the theorem, but unfortunately I don't get much out of it. The only example I sort of got was the computation of $\pi_1(\Bbb{R}^3 - S^1)$.

I would appreciate it very much if I could see some other examples to illustrate this theorem. In particular, I heard that you can use it to compute group presentations for the fundamental group - it would be good if I could see examples like that.

Thanks.

Edit: Is there a way to rigorously prove that $A_1 \cap A_2$ deformation retracts onto the point $\{p_0\}$?

share|improve this question
1  
This could be useful. Have a look at example 13.14 and 13.19. –  Matt N. Aug 5 '12 at 11:10
    
You didn't define what $W_1$ and $W_2$ are. I imagine Hatcher wants you to assume that $W_1$ and $W_2$ are both nice enough for the required deformation retract to exist. For example, if both $X_1$ and $X_2$ are manifolds, then such neighbourhoods exist because a manifold is locally euclidean, and so locally convex. –  Zhen Lin Aug 5 '12 at 11:16
    
@ZhenLin Yes I will put that in. –  user38268 Aug 5 '12 at 11:17
    
@ZhenLin How do I prove that $A_1 \cap A_2$ deformation retracts onto $p_0$? –  user38268 Aug 5 '12 at 11:18
    
@BenjaLim: You should check out Lee's "Introduction to Topological Manifolds" (2nd edition), Chapter 10, where applications of Seifert-van Kampen to the fundamental groups of CW complexes and compact surfaces are discussed in rigorous detail. –  Jesse Madnick Aug 6 '12 at 2:28

3 Answers 3

A more powerful version of the theorem is in terms of groupoids using the notion of the fundamental groupoid $\pi_1(X,A)$ on a set $A$ of base points. This theory is explained in the book "Topology and Groupoids" advertised here. For example, to capture the fundamental group of the circle from the theorem you need two base points. Since the circle is THE basic example in algebraic topology, it is a bit of an anomaly to give a theorem which does not compute this example. So one gets more powerful theorems with hardly any further complication in the proofs, which I thought was a good idea in the 1960s, when the first edition of this book was published. An MAA review of the book is given here.

The more general version of the theorem is also used in the book to give a proof of the Jordan Curve Theorem, via a nice property called the Phragmen-Brouwer Property: the circle does not have this property. (Added later: Take two distinct points $d,e$ on the circle $C$: then $C \backslash \{d\}, C \backslash \{e\}$ are both connected, but $C \backslash (\{d\} \cup \{e\})$ is not connected. The general property involves disjoint closed subspaces $D,E$.)

The groupoid approach also has advantages in considering covering spaces, and orbit spaces.

@user38268: Edit 2013: I'd just like to mention how the fundamental group(oid) of a graph is calculated from a groupoid point of view. So how do we obtain a graph? Take a disjoint union say $Q$ of directed edges. These have a set $Q_0$ of end points. Then a graph $\Gamma$ is obtained by identifying these end points in some way, i.e. by a function $f: Q_0 \to V$, say. Now the fundamental groupoid of $Q$ is the disjoint union $G$ of copies of the "unit interval groupoid" $\mathcal I$, again with object set $Q_0$. In groupoid theory we can start with $G$ and $f$ and produce a new groupoid which in T&G is written $U_f(G)$ with object set $E$, and a good universal property. On p. 343 of T&G, a deduction from the groupoid van Kampen Theorem implies that $U_f(G)$ is the fundamental groupoid of $\Gamma$ on the set of vertices. But this also the free groupoid on the graph $\Gamma$.

This may not be immediately comprehensible, but it shows how a groupoid viewpoint avoids choosing base points and a maximal tree, and simply says that the free groupoid on a graph is obtained analogously to the way the graph itself is obtained by identification of vertices, but in the category of groupoids rather in the category of graphs. See also the book by Higgins Categories and Groupoids.

The emphasis on fundamental groups suggests one would describe a railway timetable in terms of return journeys and change of starting point of the return journeys. There is a more convenient way! Higgins' book was published in 1971.

share|improve this answer

Since it happens that I am quite familiar with the corresponding chapter of Hatcher's monograph, the answers for your two questions are:

1) As I understand, $A_1 \cap A_2$ is just $W_1 \vee W_2$ and the requirement is that of being able to "paste" two deformation retracts; however, these may be incompatible (if they are arbitrary). Yet, Hatcher only cares about wedge sums of spheres (or manifolds) if he applies this example; the fact that compatible pairs of retractions do exist is enough in this case.

2) Hatcher is trying to motivate the theorem by giving a few examples with a geometrical "flavor", in which the actual deformation retracts are, in fact, difficult to write down (or even to visualize, as in "Linking Circles"). This is just one peculiarity of his writing, making it difficult at first sight. You're not the first to complain about his style, yet this work is anything but a "light read" (especially deeper into it).

You could read at first the actual theorems: given as a "proposition", 1.26 is computationally the main result of the section, helping you to do what you asked for (computing a group presentation), and also suggesting how to approach the construction of $K(G, 1)$ spaces.

Again, you may reconsider stating the elementary version (only two path-connected components) in terms of the "amalgamated" free product. At least personally, viewing the construction as a solution to a universal mapping problem (pushouts in Grp are exactly that) helped me clear a few confusions in the preamble of Hatcher's exposition about SVK.

Added: to answer the title of the topic, section 1.2 of the book is full with such examples and applications (following the actual proof), and the exercises in the end are non-trivial ones.

share|improve this answer

We use the Seifert Van-Kampen Theorem to calculate the fundamental group of a connected graph. This is Hatcher Problem 1.2.5:

It is a fact in graph theory that any connected graph $X$ contains a maximal tree $M$, namely a contractible graph that contains all the vertices of $X$. Now if the maximal tree $M = X$, then we are done because for any $x_0 \in M$, $\pi_1(M,x_0) = \pi_1(X,x_0) = 0$ that is trivially free. Now suppose $M \neq X$. Then there is an edge $e_i$ of $X$ not in $M$. Observe that for each edge $e_i$ we get a loop going in $M \cup e_i$ about some point $x_0 \in M$. Now fix out basepoint $x_0$ to be in $M$ and suppose that the edges not in $M$ are $e_1,\ldots,e_n$. Then it is clear that $$X = \bigcup_{i=1}^n \left(M \cup e_i\right).$$

The intersection of any two $M \cup e_i$ and $M \cup e_j$ contains at least $M$ and is path connected, so is the triple intersection of any 3 of these guys by the assumption that $X$ is a connected graph. So for any $x_0 \in M$, the Seifert-Van Kampen Theorem now tells us that

$$\pi_1(X,x_0) \cong \pi_1(M \cup e_1,x_0) \ast \ldots \ast \pi_1(M \cup e_n,x_0)/N$$

where $N$ is the subgroup generated by words of the form $l_{ij}(w)l_{ji}(w)^{-1}$, where $l_{ij}$ is the inclusion map from $\pi_1((M\cup e_i) \cap (M \cup e_j),x_0) = \pi_1(M \cup (e_i \cap e_j),x_0)$. Now observe that if $i \neq j$ then $M \cup (e_i \cap e_j) = M$ and since $\pi_1(M,x_0) = 0$ we conclude that any loop $w \in \pi_1(M \cup (e_i \cap e_j),x_0)$ in here is trivial. If $i = j$, $l_{ij}$ is just the identity so that our generators for $N$ are just

$$l_{ij}(w)l_{ji}(w)^{-1} = ww^{-1} = 1$$

completing our claim that $N$ was trivial. Now for each $i$, we have that $\pi_1(M\cup e_i,x_0)$ is generated by a loop that starts at $x_0$ and goes around the bounded complementary region formed by $M$ and $e_i$ and back to $x_0$ through the maximal tree. Such a path back to $M$ does not go through any other edge $e_j$ for $j$ different from $i$. It follows that $\pi_1(X,x_0)$ is a free group with basis elements consisting of loops about $x_0 \in M$ as described in the line before.

share|improve this answer
3  
If you downvoted because my answer is wrong, please tell me. How will I be able to correct the proof if you don't tell me what is wrong? –  user38268 Aug 19 '12 at 5:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.