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Hei, guys! I'm having some problem solving the next exercise:

Let $f: M -> N$ be a homeomorphism. Define a map $f*:π_1 (M, x_0) → π_1 (N, f(x_0 ))$ such that $f*([\gamma])=[f∘\gamma]$. Show that $f*$ is an isomorphism. Check that the map is well-defined.

Any suggestions?

Chan
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    Could you elaborate on what you already tried and with wich part you have problems? – jorst Oct 12 '14 at 15:11
  • I was wondering if I could use the fact that if M and N are homeomorphic, then f is a homotopy equivalence and the two fundamental groups are isomorphic. But if I have to show it explicitly, then I don't really know how to do it. – Chan Oct 12 '14 at 15:36
  • try to go through the definitions and see what has to be checked: For example for being well defined you have to show that if you take another loop $\gamma'$ homotopic to $\gamma$ via a Homotopy $H$, you have to write down a homotopy from $f\circ\gamma'$ to $f\circ\gamma$. Which one could this be? – jorst Oct 12 '14 at 15:50

1 Answers1

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We just need to prove that the fundamental group satisfies two properties below.

  • If $\text{id}:(M,x_0)\to(M,x_0)$, then $\text{id}_*=\text{id}:\pi(M,x_0)\to\pi(M,x_0)$.

  • If $f:(L,x_0)\to(M,y_0)$, $g:(M,y_0)\to(N,z_0)$, then $(g\circ f)_*=g_*\circ f_*:\pi(L,x_0)\to\pi(M,z_0)$.

So if $f$, $g=f^{-1}$ are homeomorphism, then $$\text{id}=\text{id}_*=(f\circ g)_*=f_*\circ g_*$$ $$\text{id}=\text{id}_*=(g\circ f)_*=g_*\circ f_*$$ Now we get the conclusion.

Actually the proof is based on the theory of functor: http://en.wikipedia.org/wiki/Functor.

If you want prove it by the homotopy, I shall give the next theorem.

Theorem Suppose $f$, $g:M\to N$ are continuous, and $H$ is the homotopy map between them. For any $q\in X$, let $h$ be the path in $Y$ from $f(q)$ to $g(q)$ defined by $h(t)=H(q,t)$. Now we have an isomorphism between $\pi(X,f(q))$ and $\pi(Y,g(q))$.

You can find clue in the Lee's book: Introduction to Topological Manifold, P164.

gaoxinge
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