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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?

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    Why is $F$ injective? A priori you could have 2 different points with same image. – Charlie Mar 22 '20 at 22:26
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    If I remember correctly, the definition of "embedding" for smooth manifolds requires not only that the map be one-to-one but also that its differential be everywhere one-to-one, i.e., of full rank as a linear map of tangent spaces. (This problem is in addition to the one pointed out by @Charlie and by John Hughes.) – Andreas Blass Mar 22 '20 at 23:29
  • @AndreasBlass- Yep. I think that's why the book defines a particular embedding map of the form $M^m\to (\Bbb{R}^n)^k\times \Bbb{R}^k$, where $p\to$ (each chart , value of each partition function at $p$). Given this map, which implies injectivity of the map as well as of the tangent map, does my method of dimensional reduction work? –  Mar 23 '20 at 00:09

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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.

John Hughes
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  • Ah. I see. If I include charts in the definition of $F$, maybe this proof will work? –  Mar 22 '20 at 22:32
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    I don't honestly recall the details of the original proof. I saw it once, about 47 years ago, and so I don't feel all that bad for forgetting it. I think that maybe it's in one of Milnor's books somewhere, in which case it'll be the perfect version of the proof. :) – John Hughes Mar 22 '20 at 22:36
  • No ... wait. Maybe I do recall it. Take an atlas with $k$ charts $\phi_i: U_i \to |bb R^n$, and a partition of unity $\alpha_i$ subordinate to that cover. Let $\psi_i = \phi_i \cdot \alpha_i$, which can be defined on all of $M$. Concatenate the $\psi_i$ to get a map $M \to \Bbb R^{nk}$. I think that THAT map, or something very like it, is an embedding. – John Hughes Mar 22 '20 at 22:40
  • Does my reduction of dimension technique work, given that I can construct a suitable monomorphism? –  Mar 23 '20 at 00:25
  • Thanks for your answer and helpful comments –  Mar 23 '20 at 00:25
  • The dimension-reduction is pretty much correct, but consider $h(x) = (x^{\frac13}, x)$ as a "high dimensional" embedding of $\Bbb R$ into $\Bbb R^2$. You find the vector $v = (0, 1)$ is not $h(a) - h(b)$ for any $a$ and $b$, so you project, and get $k(x) = x^\frac13$, which is not an embedding. The problems is that you need to avoid projecting along any tangent vector direction as well (to avoid a point of low-rank in $Dk$). But choosing a good direction for that goal also amounts to avoiding a $2n$-manifold, so you just avoid both of them, and you're good. – John Hughes Mar 23 '20 at 11:45
  • Thanks for that. So if I consider the closure of the embedding $M^m\times M^m\to\Bbb{R}^{n}\to\Bbb{P}^{n-1}$, I should be good, I think? –  Mar 23 '20 at 13:34
  • Assuming $M$ is closed and has no boundary, you have $M \times M \subset \Bbb R^n$ is already closed. The problem is that the limits of $f(p) - f(q)$, as $p$ approaches $q$, don't give you tangent vectors -- they give you zero. You really need something more like $\frac{f(p) - f(q)}{| f(p) - f(q) |}$ to get tangent vectors. – John Hughes Mar 23 '20 at 21:32
  • I was thinking that as $p$ approaches $q$, the tangent vectors at $q$ is approached by $f(p)-f(q)$. Hence, taking the limit would include the tangent directions in $\Bbb{P}^{n-1}$ –  Mar 23 '20 at 22:05
  • Yes, but that's not true -- as $p$ approaches $q$, a scalar multiple of the tangent vec is approached by $f(p) - f(q)$. Unfortunately, that scalar multiplier is zero. :) You might take a look at Whitney's proof of the Whitney-Graustein theorem for another application of this "look at the difference quotient over all pairs of points in a manifold" idea. – John Hughes Mar 23 '20 at 23:39
  • Going back two comments, you have $M \times M \to \Bbb R^n \to \Bbb P^{n-1}$, but that second map isn't defined on $\Bbb R^n$ --- it's defined on $\Bbb R^n - {0}$. And the "closure" you're describing will exactly fill in the $0$-point in $n$-space, which will be exactly where the second map is undefined. – John Hughes Mar 23 '20 at 23:41