Seifert-van-Kampen and free product with amalgamation

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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