Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a topological space and $f:X\to X$ be a homeomorphism. Then the induced map $f_1:\pi_1(X,x)\to\pi_1(X,fx)(\cong \pi_1(X,x))$ is an isomorphism (automorphism up to conjugate). In the following we will adapt this viewpoint and denote $f_1:\pi_1(X)\to\pi_1(X)$.

I think $f_1$ is kind of special if $f$ can be extended to another ambient space. More precisely I want to know:

  • Let $i:N\to M$ be an embedding of a submaniold $N$ into a closed manifold $M$ such that $i_1(\pi_1(N))=0$. Then for any homeomorphism $f:M\to M$ with $f(N)=N$, what can we say about the restriction $g=f|_N:N\to N$?

  • In general assume $i:N\to M$ satisfies $\ker(i_1)\neq0$ and $fN=N$. Letting $g=f|_N$, will we have $g_1(\alpha)=\alpha$ for all $\alpha\in\ker(i_1)\le\pi_1(N)$?

See here for more answers with some other interesting examples.

Thank you all!

share|cite|improve this question
In your first point: you need to demand that $f(N)\subseteq N$, for otherwise there is no such restriction. – Mariano Suárez-Alvarez Oct 21 '11 at 11:49
Oh surely I need. Thank you! – Pengfei Oct 24 '11 at 9:54
Nitpicking, $\pi_1(X)$ is not defined - you need to pick a basepoint. – Colin McQuillan Oct 24 '11 at 11:41
Thanks Colin! I will modify it. – Pengfei Oct 29 '11 at 8:59
up vote 2 down vote accepted

In both cases the answer is no; we can extend an arbitrary homeomorphism $g:N\to N$ to $Cg:CN\to CN$, where $CN$ is the cone $(N \times I)/(N \times \{0\})$, by $(Cg)(n,t)=(g(n),t)$. So your conditions don't say anything about the restriction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.