Seifert-van-Kampen and free product with amalgamation

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 $\lim\limits_{k \rightarrow \infty} k\mathbb{Z}$?

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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.

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 $\lim_{k\to\infty} k\mathbb{Z}$.

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
$$\pi_1(X)\cong\pi_1(X_1)\ast\pi_1(X_2)/\langle\!\langle [\alpha][\alpha]^{-1};\;[\alpha]\!\in\!\pi_1(X_1\!\!\cap\!\!X_2)\rangle\!\rangle.$$