I have read through this answer: Fundamental Group of Connected Sum of $n$ Tori with point removed, which deals with the special case of removing a single point. I wish to generalize this. Would anyone be able to nudge me in the right direction? Thanks.
Fundamental group of the connected sum of $n$ tori with $m$ points removed
algebraic-topologyfundamental-groupshomotopy-theory
Related Solutions
But what happens in the case of more than one diameter?
A sphere with two diameters is homotopy equivalent to the wedge sum of a sphere with a bouquet (wedge sum) of three circles. To see this:
- Start with your sphere with two diameters in its interior
- Redraw the two diameters exterior to the sphere, which is homeomorphic.
- Slide the points of contact of the diameters until they coincide, which is homotopy equivalent (so now you have the wedge sum of the sphere with two circles intersecting at opposite points)
- And finally then collapse one of the edges to a point.
As shown in the sketch below, you now have $S^2\vee S^1\vee S^1\vee S^1$. Hence the fundamental group is free group on three generators.
Similarly, three diameters will give you the wedge sum of a sphere with a bouquet of five circles. Fundamental group is free group on five generators.
Also, I want to compute the fundamental group based at point $(1,0,0)$. Does the base point matter for the fundamental group of this space or it would be same for any point in my space?
Basepoint does not matter if the space is path connected, which this is.
Your final answer is only coincidentally correct. The reasoning is wrong.
The major problem with your reasoning is that you take the homotopy equivalence $S^{n-1}\simeq\mathbb{R}^n\backslash\{0\}$. But $X\simeq Y$ doesn't imply that $X\backslash\{x_0\}\simeq Y\backslash\{y_0\}$ for some $x_0\in X$ and $y_0\in Y$. For example $X=\mathbb{R}^2$ is homotopy equivalent to $Y=\mathbb{R}$ but there is no pair of points $x_0\in X,y_0\in Y$ making $X\backslash\{x_0\}$ homotopy equivalent to $Y\backslash\{y_0\}$. In fact $X\backslash A$ is not homotopy equivalent to $Y\backslash B$ for any two nonempty finite subsets, even of different size (simply because the former is always connected while the latter is always disconnected). Even more: their fundamental groups (regardless of the choice of base points) are always different.
So you need some other topological invariant that on one hand induces the same fundamental group and on the other hand is preserved when removing points. And one such invariant is obviously homeomorphism. Therefore lets slightly modify your idea: use the stereographic projection to show that $S^{n-1}\backslash\{x_0\}\cong\mathbb{R}^{n-1}$ (note the different "$\cong$" sign, i.e. homeomorphic). And so instead of gaining points you lose them, and you get the correct formula
$$\pi_m(S^n\backslash\{x_0, x_1,\ldots, x_k\})\cong\pi_m(\mathbb{R}^n\backslash\{y_1,\ldots, y_k\})$$
(here "$\cong$" means "group isomorphism"). The relationship between $x$ and $y$ points is that $\{y_i\}_{i=1}^k$ is the image of $\{x_i\}_{i=1}^k$ under the stereographic projection based at $x_0$ (side note: with this observation it can be easily generalized to any difference $S^n\backslash A$, even when $A$ is infinite). But since $\mathbb{R}^n\backslash A$ is homeomorphic to $\mathbb{R}^n\backslash B$ for any two finite subsets $A,B$ of the same size (which I leave as an interesting exercise) then we can choose $\{y_i\}_{i=1}^k$ points arbitrarly.
Finally for $m=1$ and $n\geq 3$ the formula yields the trivial group. By Seifert-van Kampen for example, or by noticing that $\mathbb{R}^n\backslash\{y_1,\ldots, y_k\}$ is homotopy equivalent to the wedge sum of spheres of dimension $n-1$.
Related Question
- Fundamental Group of Connected Sum of $n$ Tori with point removed
- Every connected, open surface with the infinitely generated fundamental group is the interior of some non-compact surface with boundary
- Calculate the fundamental group of a Klein bottle with two points removed
- General Topology – Computing the Fundamental Group of a Quotient Space
Best Answer
Draw the polygonal representation of your n-holed torus.
Put your $k$ "missing" points in a small disk around the center of your $4n$-gon.
Draw a circle $S_1$ around the $k$ deleted points; draw a circle $S_2$ outside that.
Then the portion of the figure outside $S_1$ is topologically an $n$-holed torus with $1$ point removed, so you can use the previous computation to figure out generators and relations for its fundamental group.
The portion INSIDE $S_2$ is a disk with $k$ holes; you can compute $\pi_1$ for that as well. And then you can use Seifert-van Kampen to compute $\pi_1$ of the union of the two.