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$f:M\rightarrow N$ be a injective immersion, where $M$ and $N$ are same dimensional manifold with out boundary, we need to show $f$ is a covering map.

what I tried is, $df_x:T_x(M)\rightarrow T_{f(x)}(N)$ is injective and as $M$ and $N$ has the same dimension the map is isomorphism of vector spaces.Hence $f$ is surjective submersion also. So every point of $M$ is a regular value for $f$, now as $M$ is compact $f^{-1}(y)$ is finite, ingeneral I guess $f$ will become a proper map right? Now take any neighborhood $U$,of $y$, can I just say that $f^{-1}(U)$ is disjoint union of neighborhoods around the points $x_1,\dots,x_k$ where $f^{-1}(y)=\{x_1,\dots,x_k\}$? and $f$ maps homeomorphically those neighborhoods onot $U$? Thank you for help and correction of my answer in advance.

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Immersion implies you have an open map (inverse function theorem), and this is already enough to conclude covering space. In fact if both manifolds are connected, your map is a homeomorphism. – user641 Aug 31 '12 at 7:41

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