The Whitney Embedding Theorem says that
Any smooth compact $m$ dimensional manifold $M$ can be smoothly embedded in $R^{2m+1}$.
Bredon's "Topology and Geometry" says that the proof of this is beyond our scope. I seem to have found an extremely easy proof of this fact, and hence it has to be wrong. Could you please tell me where I'm going wrong?
Idea: Find an immersion in a higher dimensional space, using only partition functions and no charts, and then find a hyperplane onto which you can project the image injectively. The argument is similar to what is used for algebraic curves, and that is why I am suspicious that it will not work for badly behaved manifolds in general.
Proof: As $M^m$ is compact, there exist a finite number of smooth functions $\{f_1,
\dots,f_n\}$ forming a partition of unity. Consider the map $F: M^m\to \Bbb{R}^n$ defined as $p\to (f_1(p),\dots,f_n(p))$. Clearly this is a monomorphism
Now also consider the map $M^m\times M^m\to \Bbb{R}^{n}$ defined as $(p,q)\to F(p)-F(q)$. If $n>2m$, The image will be a $2m$ dimensional manifold in $\Bbb{R}^n$ (by the Rank Theorem). Hence, we can find a vector in $\Bbb{R}^n$ that has not been mapped to. Say this vector is $v$. How do we find this vector $v$? Read the edit below. Now at each point in the image of $F:M^m\to \Bbb{R}^n$, draw the line along the vector $v$, and then map each point to the intersection of this line with the plane normal to $v$. This plane would be $x.v=0$. In this way, we can reduce the dimension of the image. We can do this until we reach $\Bbb{R}^{2m+1}$. Hence proved.
EDIT: It seems clear now that to find this $v$, we will need to consider the map $M^m\times M^m\to\Bbb{R}^n\to\Bbb{P}^{n-1}$. As long as $n-1>2m$, we should be able to find such a $v$, no problem. Hence, I am not proving that $M^m$ can be embedded in $\Bbb{R}^{2m}$, but only that it can be embedded in $\Bbb{R}^{2m+1}$.
Where am I going wrong?
Best Answer
You went wrong at "clearly this is a monomorphism." For instance, on the unit circle, you can define a function $f_1$ that's positive for within 120 degrees of the north pole, and zero elsewhere; you can defined a similar function $f_2$ that's positive with 120 degees of the south pole.
And then you can make them each equal to $0.5$ for angles between $-10$ degrees and $+10$ degrees. Finally, you can replace $f_1$ with $g_1 = f_1 / (f_1 + f_2)$, and similarly for $g_2$.
Now $g_1$ and $g_2$ form a partition of unity, but on a neighborhood of the "east pole" (i.e., $\theta = 0$), both functions are constantly $\frac12$, so $\theta \mapsto (g_1(\theta), g_2(\theta))$ is not injective at all.
By the way, your proof's not far from the standard one -- the $2m+1$ case is the easy one, after all.