Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$f:M\rightarrow N$ 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.

share|cite|improve this question
    
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

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 $N$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.