Computing fundamental groups

Question

I want to prove that the fundamental group of the union of two spheres $S^m$ and $S^n$ joined to one point, and with $m,n\geq 2$ is trivial. I'm completely stuck so every help will be welcome.

Thank you in advance.

Answer 1

Your space is commonly written as $S^m\vee S^n$, and I will use that in the folowing.

There are many answers to this problem, so I'll give you two answers, using a bit of heavy machinery.

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The cellular approximation theorem

Given a map $f:X \to Y$ between two CW-complexes $X$ and $Y$, $f$ is homotopic to a cellular map, i.e. there is a $g:X \to Y$ that is homotopic to $f$ with $g(X^n) \subseteq Y^n$.

Using this theorem, and the standard CW-construction of the spheres, you get that your space consists of a $0$-cell, an $m$-cell and an $n$-cell. Any element in $\pi_1(S^m\vee S^n)$ is represented by a map $h:S^1 \to S^m\vee S^n$. Now, $S^1$ consists of a $0$-cell and a $1$-cell, so it's equal to its own $1$-skeleton.

By the theorem above, $h$ is homotopic to a map $h':S^1 \to S^m\vee S^n$ that takes $S^1$ into the $1$-skeleton of $S^m\vee S^n$. But since $m, n > 1$, the $1$-skeleton of $S^m\vee S^n$ is just the one $0$-cell. Therefore, $h$ is homotopic to a constant map, and therefore represents the identity element of $\pi_1(S^m\vee S^n)$. Since this is valid for any such $h$, we have $\pi_1(S^m\vee S^n) = 0$.

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The van Kampen theorem

The van Kampen theorem relates the fundamental group of a space with that of two connected, open subsets that cover the whole space.

Given a space $X$, and open subsets $U, V \subseteq X$ such that

• $U\cup V = X$
• $U, V$ and $U\cap V$ are path connected
• $U\cap V \neq \emptyset$

Then $$ \pi_1(X) = \pi_1(U)*_{\pi_1(U\cap V)}\pi_1(V) $$

That product notation takes some explaining. If you have two groups $G$ and $H$, then $G*H$ is called the free group generated by $G$ and $H$. It consists of all finite words of the form $s_1s_2\cdots s_n$ where $s_i \in G$ or $s_i \in H$ for each $i$. The group operation is concatenation.

Such a word can be reduced by either removing instances of the identity elements $e_G \in G$ or $e_H\in H$, or by replacing the "phrase" $g_1g_2$ with the element $g_1\cdot g_2 \in G$ (and similarily for elements in $H$). After reducing fully, we always end up with either the empty word (which is the identity element in $G*H$), or a word where every other element is from $G$ and every other element is from $H$.

The subscript to the $*$ means the following: Let $F$ be a third group, with homomorphisms $\phi:F \to G$ and $\psi: F \to H$. Let $N$ be the smallest normal subgroup of $G*H$ containing all words of the form $$ \phi(f)\psi(f)^{-1},\quad f \in F $$ Then $G*_FH = (G*H)/N$. This is called the free product with amalgation of $G$ and $H$ with respect to $F$, $\phi$ and $\psi$. In the van Kampen case, the homomorphisms $\phi$ and $\psi$ will always be the ones induced by the inclusions $U\cap V \subseteq U$ and $U \cap V \subseteq V$.

Back to our problem, let $E$ be a small, open neighbourhood around the point in $S^m\vee S^n$ where the two spheres touch. We set $X = S^m\vee S^n$, $U = S^m \cup E$ and $V = S^n\cup E$. We have that $\pi_1(U)$ is trivial, since it's homotopy equivalent to $S^m$ (it even deformation retracts onto $S_m$). For the same reason we get that $\pi_1(V)$ is trivial. Also, $U\cap V = E$ deformation retracts to a point, so it is also nullhomotopic. All in all, we get that $\pi_1(X)$ is the free product with amalgation of two trivial groups with respect to a trivial group and two trivial homomorphisms. The result is trivial.

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If you want to prove this from skratch without such heavy machinery, I'd suggest going a route close to the van Kampen theorem (in my opinion it's much easier than the variants above, but you might not always be so lucky): any loop in $S^m\vee S^n$ based at the junction point will consist of loops in $S^m$ concatenated with loops in $S_n$ in some order. Each of those parts are nullhomotopic, and therefore theo whole loop must be as well.

Answer 2

I will suggest two approaches, one using Van Kampen and other without that machinery.

First one:

I assume you have seen Van Kampen Theorem. We will refer to your space as X. Let's use the cited theorem. First we choose two open sets U,V such that $ \: U \cup V=X$. Both of them path connected and such that their intersection is path connected too, and non.empty. Let's say that U is your space with the leftmost point removed and V is your space with the rightmost point removed. It is clear that:

both are path connected and open

the intersection is not empty and it is path connected

their union is X

Now we choose the base point to be the point which you identify in both spheres.

U and V deformation retract to $S^m$ and $S^n$ respectively. So their fundamental groups are trivial as $n,m \geq 2$

The intersection of U and V deformation retract to a point (the point you use to join your spheres). Therefore the fundamental group is trivial too.

Now using Van Kampen (link to wikipedia) we deduce that the fundamental group is trivial. In fact the fundamental group is the free product of the fundamental groups of U and V with amalgamation the fundamental group of the intersection (trivial).

You can generalize this result to any wedge sum of any number of spheres using Van Kampen. And you have more examples of the power of this result here.

Second one:

Just follows from Theorem 5.12 in Armstrong, Basic Topology choosing the U,V we have chosen in the first approach. I hope you can consult that book and i encourage you to read the proof since it is really easy to understand and can be a good motivation for learning Van Kampen. Here is the statement of Theorem 5.12 in Armstrong's book:

I hope that I have helped you. If my explanation is not clear enough please ask me about any part of it.