Existence of injective function in a manifold with special atlas

Question

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!

Answer 1

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 $U_1$ .