I would like to apply Seifert van Kampen to a simple example taken from Wikipedia: I have $X = S^2$ and $A = S^2 - n$, where $n$ is the north pole and $B = S^2 - s$, where $s$ is the south pole.

According to my understanding, which might be wrong, Seifert van Kampen tells me that $\pi_1(X) = \pi_1(A) *_{\pi_1(A \cap B)} \pi_1(B)$, where the right hand side is the free product with amalgamation.

$A \cap B$, the sphere minus the two points has a fundamental group isomorphic to $\mathbb{Z}$.

The free product with amalgamation of two groups $G, H$ is $ G * H / N$, where $N$ is the smallest normal subgroup in $G * H$ (according to the Wikipedia entry about free product with amalgamation).

Note : I am aware that in the sphere example both $G$ and $H$ are trivial and so quotienting them with anything will be trivial again. This question is not about actually computing the fundamental group of $S^2$!

In the sphere example, this means I have to find the smallest normal subgroup of $\mathbb{Z}$.

Question 1: Is my understanding of Seifert van Kampen correct?

Question 2: What is the smallest normal subgroup of $\mathbb{Z}$?

As for question 2, what do you think about $\lim\limits_{k \rightarrow \infty} k\mathbb{Z}$?

Thanks for your help.

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    $\begingroup$ You have to be careful: $N$ is the smallest normal subgroup of $G \ast H$ containing $\varphi(n)\psi(n)^{-1}$, where $\varphi:N \to G$ and $\psi: N \to H$ are the inclusions. $\endgroup$
    – t.b.
    Feb 2, 2011 at 13:01
  • $\begingroup$ @Theo: Thanks! I'm not sure though what $\varphi(n)\psi(n)^{-1}$ are in my example. $\endgroup$ Feb 2, 2011 at 13:36
  • $\begingroup$ Indeed, the smallest normal subgroup of a group is always trivial... $\endgroup$
    – user1729
    Nov 14, 2011 at 13:58
  • 1
    $\begingroup$ Sadly, the question is based on a false premise: in general, it is false that $\pi_1(A) \ast_{\pi_1(A \cap B)} \pi_1(B) \cong \pi_1(X)$. To reach this conclusion one has to assume injectivity of the natural homomorphisms $\pi_1(A\cap B)\to \pi_1(A)$, $\pi_1(A\cap B)\to \pi_1(B)$. $\endgroup$ Feb 10, 2020 at 22:31

3 Answers 3


This is slightly too long for a comment, therefore I'm posting it here:

In complete generality: Let $S$ be a subset of a group $G$. Then you're familiar with the subgroup $\langle S \rangle$ generated by $S$. This is the smallest subgroup of $G$ containing $S$. Similarly, there is $N = \langle \langle S \rangle \rangle$, the smallest normal subgroup generated by $S$ (sometimes also called normal or conjugate closure). While $\langle S \rangle$ consists of precisely the words of the form $s_{i_1}^{\pm 1} \cdots s_{i_{n}}^{\pm 1}$ with $s_{i_{j}} \in S$, the smallest normal subgroup consists of the words of the form $g_{1}s_{i_1}^{\pm 1}g_{1}^{-1} \cdots g_{n}s_{i_{n}}^{\pm 1}g_{n}^{-1}$ with $g_{j} \in G$.

For example, if you're writing $G = \langle S | R \rangle$, i.e. $G$ is given in terms of a presentation with generators $S$ and relations $R$, then you actually mean $G \cong F(S) / \langle\langle R \rangle \rangle$, or in words $G$ is the quotient of the free group on $S$ modulo the normal soubgroup generated by the relations. The normal subgroup generated by a set is tremendously hard to determine (that's why the Word problem is so difficult, i.e. not solvable in general).

Now you should be able to understand what $G \ast_{A} H$ is: it's $G \ast H/ \langle\langle R \rangle \rangle$, with $R = \{\varphi(a)\psi(a)^{-1}\,:\,a \in A\}$, where $\varphi:A \to G$ and $\psi: A \to H$ are the given homomorphisms.

In your concrete situation, the situation is quite silly: As $G \ast H = \{1\}$, we must have that $N = \{1\}$.

If you want to learn about free products with amalgamation in general, the standard reference is J.-P. Serre's Arbres, amalgames et $SL_{2}$ (translated as Trees). As for Seifert-van Kampen, I think Allen Hatcher has a pretty lucid and detailed explanation on pages 40ff of his algebraic topology book, available on his home page.

  • 2
    $\begingroup$ Excellent and concise comment, clarifying much of my misconceptions and confusions. Thanks, +1. $\endgroup$
    – Leo
    May 22, 2011 at 22:01
  • $\begingroup$ @Leon: Thank you! That's nice to know, I really appreciate it. $\endgroup$
    – t.b.
    May 22, 2011 at 22:14
  • $\begingroup$ @Theo: if I see correctly, for every homomorphism $f$, we have $f(\langle S\rangle)=\langle f(S)\rangle$ and $f(\langle\langle S\rangle\rangle)\leq\langle\langle f(S)\rangle\rangle$. Is this true? $\endgroup$
    – Leo
    May 27, 2011 at 21:19
  • 1
    $\begingroup$ @Leon: All three identities you're asking about are correct . The two in the first comment follow from the fact that $f$ is a homomorphism, while the one in the second comment follows from $\langle X\,|\,R \rangle \ast \langle X' \,|\,R' \rangle = \langle X \amalg X' \,|\,R \amalg R' \rangle$ and the description on the amalgamation I gave above. Just a minor quibble: the unions in the second comment should be disjoint unions (that's why I write $\amalg$ instead of $\cup$), as a priori nothing prevents $X$ and $X$ to have elements in common. $\endgroup$
    – t.b.
    May 27, 2011 at 23:35
  • 1
    $\begingroup$ @Leon: No bother at all! What you say is perfectly correct. Note also that you can replace a relation of the form $r = abcd$ by $bcda$, $cdab$ (that is, you can cyclically permute the letters). This follows from $a^{-1}(abcd)a = bcda$ and the fact that you're dividing by the normal subgroup generated by the relations. $\endgroup$
    – t.b.
    Jun 9, 2011 at 3:17

Your understanding of Seifert van Kampen seems correct. However, note that for any group, 'the smallest normal subgroup' is of course the trivial subgroup {0}. Indeed, you left out an important part of the wikipedia sentence:

"(...) the smallest normal subgroup N of G*H containing all elements on the left-hand side of the above equation, which are tacitly being considered in G*H by means of the inclusions of G and H in their free product."

As you said, A and B have trivial fundamental groups. So in this case you are computing $\{0\}\star_{\mathbb{Z}}\{0\}$. All maps involved (i1,i2,.. in the wikipedia diagram) are therefore trivial. So this is the trivial group, as you already knew.

For a nice explanation of the amalgamated free product, with emphasis on the universal property, see The Unapologetic Mathematician.

I am not sure what you mean by $\lim_{k\to\infty} k\mathbb{Z}$.

  • $\begingroup$ @wildlife: Thanks! I gave you a vote but have not yet accepted your answer because I still don't know what $N$ is in my example. In particular, what are "the elements on the left-hand side" in my example? $\endgroup$ Feb 2, 2011 at 13:39
  • $\begingroup$ As for $lim_{k \rightarrow \infty} k \mathbb{Z}$: I was trying to find a small normal subgroup of $\mathbb{Z}$. Since $k \mathbb{Z}$ is a subgroup for all $k$ I thought, the bigger $k$, the smaller the subgroup. $\endgroup$ Feb 2, 2011 at 13:48
  • 1
    $\begingroup$ Well, what are phi and psi? They are maps from $\mathbb{Z}$ to the fundamental groups of A and B, i.e. maps $\mathbb{Z}\to \{0\}$. So both phi and psi are trivial maps, sending everything to 0. Hence the 'elements on the left-hand side' is just one element, 0 (or 1 in the notation of wiki). So N is the trivial group. Of course, there was no other option, because G*H=\{0\}*\{0\} was trivial to begin with. $\endgroup$ Feb 2, 2011 at 13:52
  • $\begingroup$ @wildlife: And thanks for the link, I'm reading it now. $\endgroup$ Feb 2, 2011 at 13:54
  • 1
    $\begingroup$ So you might want to think of the universal property (stated in that link) as the definition of the amalgamated free product. The discussion at wikipedia then gives a construction, proving that it actually exists, and how you might compute it. But in simple cases, such as this example, you only need the universal property to actually compute the am. free product: there are two trivial maps from Z to 0, and the am.free product is such that it makes the diagram commute in a universal way. This is only possible for the trivial group itself. $\endgroup$ Feb 2, 2011 at 14:05

The only thing I'd like to add is that in general, if $i:A\hookrightarrow X$ is the inclusion, then $i_\#:\pi_1(A)\rightarrow\pi_1(X)$, defined by $i_\#:[\alpha]\mapsto[i\circ\alpha]$, usually isn't an inclusion.

For example, for the inclusion $i:\mathbb{S}^1 \hookrightarrow\mathbb{B}^2$, the homomorphism $i_\#$ isn't an inclusion, since $\pi_1(\mathbb{S}^1)\cong\mathbb{Z}$ and $\pi_1(\mathbb{B}^2)\cong\{1\}$.

Even if $i$ maps $[\alpha]$ to $[\alpha]$, the first equivalency class of $\alpha$ is computed via homotopies in $A$, while the second via homotopies in $X$! The codomain of the loop is vital. However, if $A$ is a retract of $X$ (i.e. $A\!\subseteq\!X$ and $\exists$ continuous $r:X\rightarrow A$ with $r|_A=id_A$), then $i_\#:\pi_1(A)\rightarrow\pi_1(X)$ is an inclusion.

Hence in the formulation of Seifert van Kampen theorem, when one has the inclusions $i:X_1\!\cap\!X_2\hookrightarrow X_1$, $j:X_1\!\cap\!X_2\hookrightarrow X_2$, the theorem states (under the hypotheses) that $$\pi_1(X)\cong\pi_1(X_1)\ast\pi_1(X_2)/\langle\!\langle i_\#([\alpha])j_\#([\alpha])^{-1};\;[\alpha]\!\in\!\pi_1(X_1\!\!\cap\!\!X_2)\rangle\!\rangle$$ and NOT that $$\pi_1(X)\cong\pi_1(X_1)\ast\pi_1(X_2)/\langle\!\langle [\alpha][\alpha]^{-1};\;[\alpha]\!\in\!\pi_1(X_1\!\!\cap\!\!X_2)\rangle\!\rangle.$$


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