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I am trying do the following question:

Let $M$ be a $n$-dimensional smooth manifold that admits an atlas with only two charts. Show that there exists an injective smooth map $\varphi:M\to\mathbb{R}^{2(n+1)}$.

Well, I think unit partition helps, but I don't know how build a map with the earlier properties. Some hint?

Maybe the unity partition can be used to "glue" each coordinate map, but I don't know how this can be helpful.

Thanks!

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  • $\begingroup$ The key is that the manifold admits an atlas with only two charts. What does that mean? (In particular, you should write down what this means precisely.) $\endgroup$ – Adam Saltz Jun 30 '14 at 20:56
  • $\begingroup$ @AdamSaltz, I know that there exists two coordinates functions (in particular, diffeomorphism) such that the intersections is a diffeomorphism. Thus, in each chart is possible build a injective function, but I don't see how reach the solution $\endgroup$ – DiegoMath Jun 30 '14 at 21:12
  • $\begingroup$ @AdamSaltz, do you have any hint? $\endgroup$ – DiegoMath Jun 30 '14 at 21:12
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    $\begingroup$ @Adam: the meaning of an atlas having 2 charts is perfectly clear. If you know how to solve the problem, sketch a proof. Asking the OP to write down the definitions of concepts appearing in his question is rude and useless. $\endgroup$ – Georges Elencwajg Jun 30 '14 at 21:23
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Let $f_i:U_i\to V_i\subset \mathbb R^n \;(i=1,2)$ be the two charts and let $\rho_i\;(i=1,2)$ be a partition of unity subordinate to the covering $(U_i)$.
The required map can then be taken as

$$\phi=(\rho_1f_1,\rho_2f_2,\rho_1, \rho_2):M\hookrightarrow \mathbb R^n\times \mathbb R^n \times \mathbb R \times \mathbb R=\mathbb{R}^{2(n+1)}$$

[As usual $\rho_if_i:M\to \mathbb R^n$ is defined to be $0\in \mathbb R^n$ outside of $U_i$]

Injectivity of $\phi$ is shown as follows:
Suppose $\phi(m)=\phi(p)$.
Since $\rho_1(m)+\rho_2(m)=1$, we must have $\rho_1(m)\gt0$ (say). But then $\rho_1(p)=\rho_1(m)\gt0$ too, which forces $m$ and $p$ to both be in $U_1$ and thus $m=p$ by the injectivity of $f_1$.

Note carefully
The only way for a point $x\in M$ to satisfy $\rho_1(x)\gt0$ is to be in $U_1$.
But even this is not sufficient: there exist points $y\in U_1$ with $\rho_1(y)=0$ : intuitively they are the points $y$ near the boundary $\partial U_1$ of $U_1$ .

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  • $\begingroup$ Why $\rho_2(m)=\rho_1(p)$? I think you mean say $\rho_2(m)=\rho_2(p)$. Can you explain? $\endgroup$ – DiegoMath Jun 30 '14 at 22:29
  • $\begingroup$ Dear Diego, that was an unfortunate typo, now corrected: thanks for catching it. $\endgroup$ – Georges Elencwajg Jun 30 '14 at 22:40
  • $\begingroup$ No problem. By the way, beautiful solution. $\endgroup$ – DiegoMath Jun 30 '14 at 22:43
  • $\begingroup$ A last question: Is this function well defined in the overlaps? $\endgroup$ – DiegoMath Jun 30 '14 at 22:48
  • $\begingroup$ Dear Diego: yes the function is well defined everywhere with the convention alredy mentioned that $\rho_if_i:M\to \mathbb R^n$ is defined to be $0\in \mathbb R^n$ outside of $U_i$ $\endgroup$ – Georges Elencwajg Jun 30 '14 at 22:50

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