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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!

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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

1 Answer 1

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.

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