I want to prove that no finite group acts free in $ \mathbb{R}^n $ in the process I found the following doubt:
How can you prove that: If $ E \rightarrow X $ is a covering, then $ |\pi_1 (X)|$ divides to $ \chi (E) $?
2026-03-26 11:04:28.1774523068
If $ E \rightarrow X $ is a covering, then $ |\pi_1 (X)|$ divides to $ \chi (E) $
46 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in ALGEBRAIC-TOPOLOGY
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