# $\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+1} -A) \Rightarrow \pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+k} -A) \ (k \geq 1)$?

Let $A$ be a subspace of $\mathbb{R}^{n-1}$ with $n \geq 4$. And if we know $\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+1} -A)$, then can we say that $\pi_1(\mathbb{R}^n -A) \cong \pi_1(\mathbb{R}^{n+k} -A)$ for all $k \geq 1, k \in \mathbb{Z}$ ? I have tried to prove it or find an counter example, but failed. Any suggestions and ideas are appreciated.

Of course, we are given that $\mathbb{R}^n -A$ is path-connected.

Sorry, I shall be more specific. Here $n$ is a fixed number. And we may want to do induction on $k$. I am sorry for any confusion.

And I am not sure whether this statement is correct or not. This problem arose while my friends and I were discussing a homework problem.

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There is no need to invoke the axiom of choice to conclude that isomorphism is transitive... @user5980: Are you sure you copied the statement correctly? It is hard to believe the answer is "isomorphism is transitive"! –  Mariano Suárez-Alvarez Mar 16 '11 at 17:35
My thought would be that the fundamental group must be trivial in order for this to hold. –  JSchlather Mar 16 '11 at 22:03
@InterestedQuest: I think that user5980 is assuming that that the statement holds for a given $n$, not all values of $n$, and from that trying to prove by induction that the condition holds for $n+k$. –  Alex Becker Mar 16 '11 at 22:44
@Mariano Suárez-Alvarez: I am not sure about my problem. We "invented" this problem during discussion. @Alex: Yes. This is what I mean. –  Junyu Mar 17 '11 at 12:44
If $A$ is a subspace of $\mathbb{R}^{n-1}$, what do you mean by $\mathbb{R}^n - A$? Do you mean $\mathbb{R} \times (\mathbb{R}^{n-1} - A)$? Or do you mean to regard $\mathbb{R}^{n-1} \subset \mathbb{R}^n$? In the first case, the fundamental group of a product is the direct product of fundamental groups. In the second case, have you tried using Van Kampen's theorem? –  Robert Bell Mar 17 '11 at 13:00
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Well, I think Jacob is right.

First note that all spaces involved in the statement are connected.

We need 2 cases:

a) $A=\mathbb{R}^{n-1}$. In this case $\pi_1(\mathbb{R}^n -A) \neq \pi_1(\mathbb{R}^{n+1} -A)$

b) $A \neq \mathbb{R}^{n-1}$. Then we claim $\pi_1(\mathbb{R}^{n+1} -A)=0$ (this claim fairly obviously implies $\pi_1(\mathbb{R}^{n+k} -A)=0$ for all $k>0$, and hence finishes the proof, as all groups in the statement are then $0$).

Proof of claim: Let $\gamma$ be a loop in $B=\mathbb{R}^{n+1} -A$.

Subclaim1: $\gamma$ is homotopic in $B$ to a loop in $C=\mathbb{R}^{n+1} -\mathbb{R}^{n-1}$.

Proof of subclaim: Project $\gamma$ along the $\mathbb{R}^{n-1}$ to $\delta$ in $\mathbb{R}^2$. Let the image of $\mathbb{R}^{n-1}$ under projection be point $p$. Then $\delta$ is homotopic to a curve missing $p$, and the homotopy can be made disjoint from $p$ at all $t>0$ (say, like in page 35 Hatcher). Then taking a product with identity homotopy to lift back to $C$ in fact gives a homotopy of $\gamma$ in $B$ to a curve $\gamma'$ in $C$.

Subclaim 2: Any loop in $C$ is contractible in $B$. Proof: Any loop in $C$ is homotopic to a meridianal one (going around the $\mathbb{R}^{n-1}$, constant in those $n-1$ coordinates). Make that a meridianal point over a point $q$ in $\mathbb{R}^{n-1}$ not in $A$. Contract it in the complementary $\mathbb{R}^2$. So it is homotopic to a constant loop. QED.

So, any curve in $B$ is contractible. Claim is proved.

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