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I want to prove that if a homeomorphism (a continuous bijection with continuous inverse) in a two dimensional manifold doesn't have a periodic point, then the Euler Characteristc of the manifold is zero.

In other words, if $M$ is a two dimensional compact manifold, and there exist a homeomorphism $h: M\longrightarrow M$ such that $h$ doesn't have a periodic point, then $\chi(M)=0$.

It implies for instance that a homeomorphism on a sphere always have periodic points.

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    $\begingroup$ What do you mean by a periodic point? If $\tau: S^2 \to S^2$ is the antipode map on the sphere, then $\tau^2(x) = x$ for all $x \in S^2$, so wouldn't this mean that $\tau$ has periodic points? $\endgroup$ – JHF Dec 7 '18 at 17:50
  • $\begingroup$ Yes, it has a periodic point. A periodic point for $h$ is a $x$ in the domain such that there exist an $n\in \mathbb{N}$ with $h^{n}(x)=x$. $\endgroup$ – Gil Astudillo Dec 7 '18 at 17:54
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    $\begingroup$ The comment of @JHF refers, perhaps, to your final sentence. Did you mean to say that a homeomorphism on a sphere always has periodic points? $\endgroup$ – Lee Mosher Dec 7 '18 at 19:39
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    $\begingroup$ One could apply the Lefschtez fixed point theorem to this situation. Assuming that $h$ preserves orientation, one could conclude that if $T \in \text{SL}_n(\mathbb Z)$ is a matrix representing the induced homology homomorphism $f_* : H_1(M;\mathbb Z) \to H_1(M;\mathbb Z)$, then $\text{trace}(T^k)=2$ for all $k \ge 1$. This must have strong consequences... for instance if it implied that $T$ is the identity then you would be done. $\endgroup$ – Lee Mosher Dec 7 '18 at 19:51
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    $\begingroup$ As a follow-up to Lee Mosher's argument, the condition on the traces of the powers implies that there are exactly two nonzero eigenvalues for $T$, both equal to $1$. Since $T$ is invertible, this means that the rank of $H_1$ is $2$, so that the Euler characteristic is zero. $\endgroup$ – JHF Dec 7 '18 at 20:19

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