Let M be a manifold and $B_p(1,M)$ a ball of radius 1 and center p in M. Let $F:B_p(1,M)\to \mathbb{R}^n$ a map such that $F(\partial B_p(1,M))\subset B_0(1+\epsilon,\mathbb{R}^n)\setminus B_0(1-\epsilon,\mathbb{R}^n)$ where $B_0(1,\mathbb{R}^n)$ is the ball in $\mathbb{R}^n$ centered at zero with radius 1 and $\epsilon$ a small number. (here $\partial B_p(1,M)$ is the boundary of $B_p(1,M)$)

Let $F_\ast:H_n( B_p(1,M),\partial B_p(1,M),\mathbb{Z}_2)\to H_n(B_0(1+\epsilon,\mathbb{R}^n),B_0(1-\epsilon,\mathbb{R}^n),\mathbb{Z}_2)$.

Suppose now that there exists $x\in B_p(\frac{1}{2},M)$ such that $F^{-1}(F(x))=x$ i.e. $F(x)$ has exactly one preimage. Why this implies that $F_\ast$ is non-zero ?

thank you !

  • $\begingroup$ I have to add that $F$ preserves essentially the boundary. $\endgroup$ – Cauchy Apr 19 '13 at 21:36
  • 2
    $\begingroup$ Cauchy: Right now, the question makes no sense, since $F_*$ is not well-defined, as $F$ is not a map of pairs. In any case, the (properly corrected) question is more suitable for math.stackexchange. $\endgroup$ – Misha Apr 19 '13 at 22:42

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