# Fundamental group of the sphere with n-points identified

I am trying to compute the fundamental group of the sphere with n points identified by using Seifert Van Kampen. I split up my surface into $U =$ the top half of the sphere including the n points and $V =$ the lower half of the sphere (not including the points), where $U \cap V$ is a cylinder.

$U$ turns out to be homotopy equivalent to the the wedge of n petals, so $\pi_1(U)$= free group on $n-1$ generators, $\pi_1(V) = \{1\}$ ( as $V$ is a disk) and $\pi_1(U \cap V) = \pi_1(S^1) = \mathbb{Z}$.

I am having trouble applying van Kampen to find the relators of $U \cup V$.

Hint: The inclusion maps $$i : U \cap V \to U \quad\text{and}\quad j : U \cap V \to V$$ induce group homomorphisms $$i_* : \mathbb{Z} = \pi_1(U \cap V) \to \pi_1(U) = F_{n-1}$$ and $$j_* : \mathbb{Z} = \pi_1(U \cap V) \to \pi_1(V) = \{1\}$$ You must compute these homomorphisms, as Van Kampen's Theorem suggests.

The second homomorphism $$j_*$$ is obviously the trivial homomorphism.

The first homomorphism $$i_*$$ is also trivial. To see why, notice that the map $$i$$ factors as $$U \cap V \mapsto \text{the boundary circle of the top hemisphere} \hookrightarrow \text{the top hemisphere} \mapsto U$$ where the first map of this sequence is a deformation retraction of the cylinder $$U \cap V$$ onto its central circle (which is the boundary circle of the top hemisphere), and last map is the quotient map under which the $$n$$ points are identified to a single point. Since the top hemisphere is simply connected, it follows that $$i_*$$ is the trivial homomorphism. (Thanks to @BennyZack for this observation in the comments.)

• I know that the second homomorphism is trivial, but Im not sure how to compute the first one. I'm used to finding the relators "visually" but in this case it seems more difficult to do it that way, but I am not sure how else to do it – bmmcutet12 Aug 5 '17 at 16:25
• I added more to my answer regarding your visualization question. – Lee Mosher Aug 5 '17 at 22:39
• I might be wrong, but isn't the second homomorphism clearly trivial since the equator is nullhomotopic within the upper upper hemisphere and this homotopy can be composed with the gluing map to get that it is also nullhomotopic in $\pi_1 (U)$? – Benny Zack Dec 5 '18 at 10:49
• Yes, that's absolutely correct, I overlooked that. It definitely simplifies things a bit, in that it's not necessary to bother trying to visualize how the circle $U \cap V$ homotopes into the wedge. I'll rewrite my answer. Thanks. – Lee Mosher Dec 5 '18 at 15:18

Extended hint:

You might want to try a slightly different approach: pick, as $U$, a neighborhood of the point $P$ to which all $n$ points are identified. Deleting a slightly shrunk version of $U$ leaves you with $V$, a sphere with $n$ holes, or, enlarging one of them a good deal, a disk with $n-1$ holes.

The problem with this is that the intersection, $U \cap V$, is not connected, alas, so Seifert-van Kampen doesn't apply directly (or you have to do something with multiple basepoints and groupoids, etc., which is way more than seems necessary). But it's a natural place to start, surrounding the "interesting" point with one of the two open sets to be used in S-vK. You just have to adjust the approach a little:

If you draw arcs from $Q_1$ to $Q_2$, $Q_2$ to $Q_3, \ldots, Q_{n-1}$ to $Q_n$, where the $Q_i$ are the points to be identifed, and then let $U_0$ be a neighborhood of this collection of arcs in the sphere, and $V_0$ be a slight expansion of $S^2 - U_0$, and then project $U_0$ into the identification space, getting a neighborhood $U$ of the "identification point" $P$, and similarly get $V$ ... then you can have a connected intersection $U \cap V$ --- it's just a circle. (I find it's helpful to place $Q_1, \ldots, Q_n$ in a small sequence on the equator of $S^2$, so that $U_0$ is a neighborhood of a small equatorial arc. The intersection of $U$ and $V$ is then just a curve that surrounds this arc.

Now you should have enough to go on. You have to look at $U_0$ and see if it deformation retracts onto something nice and simple, and you have to look at a generator for the circle $U\cap V$ and see what it becomes in $\pi_1(U)$ and in $\pi_1(V)$.

To answer questions like this one really needs to advance into the world of not necessarily connected spaces, such as a finite set of points $C$ on a sphere $X$. Then if $f: C \to Y$ is a map to a space $Y$ one can consider the space $Z= Y \cup _f X$ obtained by identifying some of the points $C$ of $X$ to points of $Y$, which we assume connected, although that can be dealt with.

The idea is to model the geometry by algebra, and this can be done by the algebra of groupoids, and the topogical construction of the fundamental groupoid $\pi_1(X,C)$ on a set $C chosen according to the geometry. Suppose now that$Y$is a discrete space$D$, so that$f: C\to D$. Let$G=\pi_1(X,C)$. Then we can construct a new groupoid written$U_f(G) $or$f_*(G)$with object set$D$, and morphism$f':G \to U_f(G) $, and having the property that the following diagram of groupoids is a pushout: $$\begin{matrix} C & \xrightarrow{f} & D\\ \downarrow & & \downarrow\\ G &\xrightarrow{f'} & U_f(G) \end{matrix}$$ where the vertical arrows are just inclusions as sets of identities. This construction, which includes those of free products of groups, and of free groups, and indeed of free groupoids on a directed graph, is due to Higgins; it is given in his book Categories and Groupoids and also in Topology and Groupoids. It is also useful to develop a number of properties of this construction: for example if$G$is connected and has trivial vertex groups, then$U_f(G)\$ is a free groupoid, and so its vertex groups are free groups.

You also have to generalise the standard van Kampen Theorem from the case of a single base point to the case of a set of base points, as was published in 1967, and so gave a theorem which could compute the fundamental group of the circle, a rather basic example in topology. This general theorem is given in the above two books.

For further discussion see this mathoverflow question.