Let $X \to Y$ be a covering with finite number of sheets, where $X$ and $Y$ are connected simplicial complexes. Can you provide an example when homology groups of $X$ and $Y$ with coefficients in $\mathbb{Z}_2$ are non-isomorphic?
Example of covering space with non-isomorphic to base space homology groups
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algebraic-topology
homology-cohomology
covering-spaces
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2Did you try surfaces of finite genus? – 2017-01-10
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1$S^2 \to \mathbb RP^2$ should be ok right ? – 2017-01-10
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0Pedro Tamaroff, thank you for such a great hint! – 2017-01-10
2 Answers
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Almost any example will do. For instance, if $Y$ is a wedge of two circles then it has a 2-sheeted covering space $X$ that is homotopy equivalent to a wedge of three circles. (If you want a picture of how this works, see example (1) on page 58 of Hatcher's Algebraic Topology.) Then $H_1(X;\mathbb{Z}_2)\cong\mathbb{Z}_2^3$ but $H_1(Y;\mathbb{Z}_2)\cong\mathbb{Z}_2^2$.
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Trying to answer my own question after receiving some useful hints.
Let $\Sigma_g$ be a surface of genus $g$, then consider a covering $\Sigma_m \to \Sigma_g$, where $m = n(g-1)+1$ and $n$ is a number of sheets. The first homology groups are $H_1 (\Sigma_m) = \mathbb{Z}_2^{2m}$, $H_1 (\Sigma_g) = \mathbb{Z}_2^{2g}$ and they are non-isomorphic.