Injective application on compact manifold and some Euclidean Space. I would like to tips for solving this problem.
Let $M^{n}$ a compact manifold. Prove that there exists $m \in \mathbb{N}$ and an injective application $\phi:M^{n}\to \mathbb{R}^{m.(n+1)}$.
Note that , if there exists $\phi$.


*

*$\phi$ is an injective immersion defined on compact set.(by rank theorem)

*$\phi$ is an smooth embedding.(injective immersions on compact sets are embeddings)

*$\phi$ is a closed map, because if $A\subset M$ is a closed set then $A$ is compact and $f(A)\subset \mathbb{R}^{m.(n+1)}$ is compact, but compact subsets of Hausdorff spaces are closed.

*$\phi$ is an open map, because $\phi$ is a homeomorphism.

*$f(M)=\mathbb{R}^{m.(n+1)}$, by (3) and (4) and because $\mathbb{R^{m.(n+1)}}$ is connected.


But,I could not finish anything. 
 A: Here are some comments that might help you get on the right track.  First off, the word you want in English is "map," not "application." "Application" is an English word, but it means something else entirely. 


*

*Your original statement of the problem only asked for $\phi$ to be an injective map. Don't you want it to be smooth? Or at least continuous? Even assuming it's smooth will not imply that it's an immersion. Once you produce the map, if you want it to be an immersion, you'll have to prove that separately.

*Yes, if you can produce a map that's an injective immersion, then it will be a smooth embedding. 

*Right, as long as $\phi$ is continuous, it will be a closed map. 

*Wrong. $\phi$ will be a homeomorphism onto its image, which means that it's open when considered as a map from $M$ to $\phi(M)$. But it won't be open as a map into $\mathbb R^{m(n+1)}$.

*$\phi$ certainly cannot be surjective if it's continuous, because $M$ is compact while $\mathbb R^{m(n+1)}$ is not.


Once you get your definitions straightened out, here's a hint: coordinate charts and partitions of unity are going to be helpful.
A: Here is a somewhat better result:
Theorem (Whitney): Any compact $n$-manifold may be embedded in $\mathbb{R}^{2n+1}$.
The proof is not so difficult, have a look at page $22$ here. It uses "Baby Sard".
But of course one can first prove an "easier" result as above. On the other hand, some ideas from above might be useful there, too.
