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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
up vote 1 down vote accepted

Since $f$ is an immersion it is a local homeomorphism. It is known (and not hard to show) that a local homeomorphism with the same non-zero number of elements in all fibers is a covering map. So we prove fibers are constant of the same cardinality.

Let $y\in N$. Then $f^{-1}(\{y\} )$ is finite as you have said. We prove that for each $y\in N$ locally around $y$ the number of elements in fibers is constant. Let $y_1,\dots ,y_m$ be all elements mapped to $y$ and let $U_i$ be open disjoint containing $y_i$. We prove there exists $V$ an open neighbourhood of $y$ such that $f^{-1}(V)\subset\bigcup U_i:=U$. Suppose this is not true, let $V_i$ be countable local basis around $y$ and let $z_i\in f^{-1}(V_i)\setminus U$. Then $f(z_i)\rightarrow y$. Let $z$ be an accumulation point of $z_i$, then $f(z)=y$ and hence $z=y_j$ for some $j$. But that means for $n$ sufficiently big $z_n\in U_j\subset U$, a contradiction. Now we decrease $U_i$ such that $f|U_i:U_i\rightarrow f(U_i)$ is diffeomorphism ($f$ is an immersion)and let $V$ be as in the proved assertion. Then cardinality of fibers for $x\in V$ are constant. Therefore cardinality of fibers are same on the connected component of $x$, which assume to be the whole of $N$.

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