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It is a brilliant idea to classify topological spaces with respect to their fundamental group. Here are some examples, partly from $\S59$ of Munkres' book.

$\pi_1(X)\cong\mathbb{Z}:$ the solid torus $D^2\times S^1$; the cylinder $S^1\times I$; the infinite cylinder $S^1\times \mathbb{R}$; the sphere $S^2$ with two points removed...

$\pi_1(X)$ isomorphic to the fundamental group of digit '8': the torus $T=S^1\times S^1$ with one point removed; $\mathbb{R}^3$ with two disjoint lines removed; $S^1\cup (\mathbb{R}\times 0)$...

The cases above are easy to imagine in $\mathbb{R}^3$, so intuitively I can solve them by looking for deformation retracts. However, I'm stuck with some more difficult cases below:

  1. $P^2$ with one point removed; ($P^2$ stands for the projection space)
  2. $P^2$ with two points removed;
  3. $S^3$ with two points removed;($S^3:=\{(x_1,x_2,x_3,x_4)\in \mathbb{R}^4|x_1^2+x_2^2+x_3^2+x_4^2=1\}$)
  4. Klein Bottle with one point removed;
  5. $\mathbb{R}^3-S^1$;
  6. $\mathbb{R}^3-S^1-l$, $l$ is the line perpendicular to $S^1$ and going through the center of $S^1$;

For 5 and 6, I guess the fundamental groups are $\mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}$ respectively. This guess carries from the idea of $\pi_1(S^1) = \mathbb{Z}$ which indicating the winding number of closed path. Is this guess correct?

For 1, 2, 3 and 4, it is hard to imagine them in $\mathbb{R}^3$, difficult to reduce them into some trivial spaces like a circle or something. Could anyone hint me?

Thank you in advance!

Links to related pages:

  1. fundamental group of the Klein bottle minus a point
atlantic0cean
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    these are all `typical' spaces to calculate the fundamental of, so you should be able to find each with a detailed explanation somewhere on math.stackexchange, or else in standard reference books such as Hatcher's Algebraic Topology. – Dan Rust Jun 28 '21 at 14:05
  • For the first one, think of $\mathbb{P}^2$ in terms of a gluing of the edges of a square, remove the center of the square, and deform the space to the boundary of the square (keeping the identifications). – Michael Burr Jun 28 '21 at 14:05
  • Hint: $RP^2$ with one point removed is the "open" Moebius band. – Moishe Kohan Jun 28 '21 at 14:43
  • @MoisheKohan What does an open Mobius band mean? – atlantic0cean Jun 29 '21 at 10:49
  • @DanRust Thanks! I've found some helpful posts on math.stackexchange and added links to my post. – atlantic0cean Jun 29 '21 at 11:15
  • For 3) you can do the following: S^3 with one point removed is homeomorphic to R^3, and R^3 with one point removed is still contractible. – amnesiac Jun 29 '21 at 11:42
  • @amnesiac Good idea! Does it hold that $S^n$ with one point removed is homeoporphic to $\mathbb{R}^n$? Another question is, generally speaking, if $A$ is homeomorphic to $B$, is $A$ with $m$ point(s) removed homeomorphic to $B$ with $m$ points removed? – atlantic0cean Jun 29 '21 at 11:47
  • @atlantic0cean The answer is yes to both questions: for the first one, look for “stereographic projections”. The second one is a straight-forward result in General Topology! (if $f:A\to B$ is a homeomorphism, then $f:A\backslash{ p} \to B\backslash {f(p)}$ is a homeomorphism as well!) – amnesiac Jun 29 '21 at 11:57
  • @amnesiac Thanks! I think what I want to ask in question 2 is: if $A$ is homotopically equivalent to $B$, then how about $A$ with a point removed and $B$ with a point removed? (I'n not a native English speaker so I didn't realize I was talking about homeomorphism instead of homotopic equivalence, so sorry for the confusion....) – atlantic0cean Jun 29 '21 at 12:14
  • @atlantic0cean In that case, no! For instance, $\mathbb{R}$ and $\mathbb{R}^2$ are both contractible (so homotopic equivalent) but can you see what happens when you remove a point in this case? – amnesiac Jun 29 '21 at 12:24
  • @amnesiac Get it! Thanks a lot! :) – atlantic0cean Jun 29 '21 at 12:28
  • This means Moebius band with boundary removed. – Moishe Kohan Jun 29 '21 at 13:15
  • @MoisheKohan Then it implies $\pi_1(P^2-{x_0})=\mathbb{Z}$, right? – atlantic0cean Jun 29 '21 at 13:19
  • @atlantic0cean: Of course. – Moishe Kohan Jun 29 '21 at 13:30

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