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?
Covering spaces and Euler characteristc obstruction
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algebraic-topology
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0I 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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1If 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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0@Adam: Is there a simple proof of that? – 2012-01-17
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0@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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0@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