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Let $X$ and $Y$ be surfaces. If we have a covering $X\to Y$, then $\chi(Y)\mid \chi(X)$. I assume that the condition on Euler characteristics is not enough to guarantee that a surface covers another. Are there any trivial examples of a surface $X$ that does not cover the surface $Y$ even though the Euler characteristic is a multiple?

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    I deleted my answer, since you want an example with nonzero Euler characteristic. You may want to state the condition in your question.2012-01-17
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    If you are fine with non-orientable examples, then each $\#_g T^2$ and $\#_{2g} \mathbb{R}P^2$ have the same Euler characteristic. However the two are distinct. i.e. there is no degree 1 cover.2012-01-17
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    @Adam: Is there a simple proof of that?2012-01-17
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    @Adam: Sorry, for the stupid question. By the classification of coverings a degree 1 cover would be the space itself.2012-01-17
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    @pji That's the point I was making: If they have the same Euler characteristic then any cover would be degree one. A degree one cover is a homeomorphism. One the other hand, we know no such homeomorphism exists since the two differ in (for example) orientability.2012-01-19

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Take $X=S^2$, the sphere. Then $\chi(X)=2\neq 0$. $Y$ to be the compact Riemann surface with genus $g(Y)=2$, which implies that $\chi(Y)=2-2g(Y)=-2\neq 0$ Clearly, $\chi(Y)=-2|2=χ(X)$. But $X$ does not cover the surface $Y$, since by Uniformization Theorem, the universal covering of $Y$ is the unit ball.