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I know there are other questions similar to this one, but I just want you to tell me if what I'm doing is rigth and how to improve it. The problem is the following: (I'm using the definitions by Guillemin & Pollack)

Let $Y\subset R^N$ be an $l-$dimensional manifold and suppose $X$ is a $k-$dimensional submanifold of $Y$. Let $\imath:X\rightarrow Y$ be the inclusion map. Then for every $x\in X$, the differential $d\imath_x:T_xX\rightarrow T_xY$ is also the inclusion map.

This is my solution: Let's take $x\in X$. Consider a parametrization $\phi:U\subset R^l\rightarrow V$ of $Y$ around $x$ such that $0\in U$ and $\phi(0)=x$. Since the definitions don't depend on the parametrizations we can consider the following parametrization of $X$ around $x$: We know that there exists an open set $\mathcal V$ in $R^N$ such that $V=\mathcal V\cap Y$, so we can consider the open set $\tilde V=\mathcal V\cap X$ in $X$. Define $\tilde\phi:\tilde U\subset R^k\rightarrow\tilde V$ as $$\tilde\phi(a_1,\dots,a_k)=\phi(a_1,\dots,a_k,0,\dots,0),$$ where $\tilde U$ is the inverse image of $U$ under the canonical inclusion $\jmath:R^k\hookrightarrow R^l$. Clearly, $\tilde\phi$ is smooth, because $\tilde\phi=\phi\circ\jmath$. Moreover, $\tilde\phi$ is a diffeomorphism with inverse $\tilde\phi^{-1}=\pi\circ\phi\vert_{\tilde V}^{-1}$, where $\pi:R^k\times R^{l-k}\rightarrow R^k$ is the projection on $R^k$. Now, let $v\in T_xX$, then $v=d\tilde\phi_0(a)$ with $a\in R^k$. Observe that $v=d\phi_0(\jmath(a))$. Then $$d\imath_x(v)=d\phi_0\circ d\tilde\imath_0\circ(d\tilde\phi_0)^{-1}(v),$$ where $\tilde\imath=\phi^{-1}\circ\imath\circ\tilde\phi=\jmath\vert_{\tilde U}$. Since $d\tilde\imath_0=\jmath$, then we have $$d\imath_x(v)=d\phi_0\circ d\tilde\imath_0\circ(d\tilde\phi_0)^{-1}(v)=d\phi_0(\jmath(a))=v.$$ So, $d\imath_x$ is the inclusion map.

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    $\begingroup$ This seems right to me. I think you could summarize the part where you choose the parametrizations by citing the Rank Theorem. $\endgroup$ – Sak Jul 9 '14 at 0:51
  • $\begingroup$ This question is pretty old by now. However, how do you know that your definition of $\tilde{\phi}$ takes $\tilde{U}$ into $\tilde{V}$? $\endgroup$ – fourierwho Jun 29 '16 at 16:37
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This is certainly a valid proof, and whilst the method behind it is very intuitive, actually writing out what's being doing is very messy, especially considering the problem feels like it should be "obvious".

A much "nicer" proof comes from using the following equivalent definition of derivative, if $f:X\rightarrow Y$ is a smooth map of manifolds embedded in $\mathbb{R}^N,\mathbb{R}^M$ respectively, then we define the derivative of $f$ at $x$, $df_x:T_xX\rightarrow T_{f(x)}Y$ by taking a smooth extension of $f$, $F:U\rightarrow\mathbb{R}^M$ (where $U$ is an open subset of $\mathbb{R}^N$) and taking $df_x$ to be the restriction of $dF_x$ to $T_xX$.

You can show that this is (well) defined (i.e. a map to $T_{f(x)}Y$, independent of choice of extension) by showing that for any given extension and choice of parameterisations $\phi$ on $X$ and $\psi$ on $Y$, $dF=(d\psi)(d\bar f)(d\phi)^{-1}$ (on suitable domains). Note that the left hand side is independent of chart, and the right hand side is independent of extension, so both expressions are independent of both! I count at least three birds with one short proof (which is one reason that I much prefer this definition of derivative) because simply showing this definition is well defined also shows that your definition is well defined, and that the two are the same, providing you multiple methods for computation.

Anyway, using this definition, with the notation as in your question, the inclusion map $\iota :X \rightarrow Y$ is just the restriction of the identity map $I_N$ on $\mathbb{R}^N$. Thus $d\iota_x$ is the restriction of the derivative of $I_N$ to $T_xX$, i.e. the inclusion map. Done in barely three lines, with no computation!

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    $\begingroup$ Of course, one way to prove that there is a local smooth extension of a smooth map on a submanifold is to apply the local immersion theorem, which tells us that there's a chart on $Y$ in which $X$ appears as a slice, just as in the OP's discussion. $\endgroup$ – Ted Shifrin Jul 9 '14 at 13:06
  • $\begingroup$ @TedShifrin Sure, but I haven't used anything like that in the above, just the definition of a smooth map on an embedded manifold. Am I missing something? $\endgroup$ – Tom Oldfield Jul 9 '14 at 13:28
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    $\begingroup$ I guess I'm trying to complete the logical circle of justifying why the two definitions are equivalent. Of course, Guillemin & Pollack first define smooth maps on arbitrary subsets of $\Bbb R^n$ in terms of a local smooth extension, anyhow. $\endgroup$ – Ted Shifrin Jul 9 '14 at 13:34
  • $\begingroup$ @TedShifrin Oh I see, yes, that's a valid point. If you're using the abstract definition of smooth, rather than the embedded one then there is more work to be done, i.e. showing that abstract smooth $\implies$ embedded smooth, to allow smooth extensions. Yes, thinking about it using a canonical inclusion map is the most obvious way to prove this. Thanks! I guess it just depends on what combinations of abstract/embedded definitions you use. $\endgroup$ – Tom Oldfield Jul 9 '14 at 13:47
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Here's a "counterexample" to the original post as requested. We'll define spaces $X \subset Y$ and a parametrization $\phi $ so that the suggested function $\tilde \phi $ is not well defined. The counterexample has a lot of pieces so I suggest making some sketches.

Let the ambient space be ${\rm I\!R}^3$. Let $Y = S^2 $ be the sphere in ${\rm I\!R}^3$ and let $$X = \left\{ {\left( {x,y,{1 \over {\sqrt 2 }}} \right)\,\,\left| {\,x^2 + y^2 = {1 \over 2}} \right.} \right\}$$

be the $45^0 $ N latitude line on the $S^2 $ globe. Let $x = \left( {{1 \over 2},{1 \over 2},{1 \over {\sqrt 2 }}} \right) \in X$ . Let $\phi :U \to V$ where $U = \{(x,y)\,|\, x^2+y^2<1/4\}$ is the open disc of radius 1/4 centered at $0$ in the x-y plane, and $$\phi \left( {x,y} \right) = \left( {x + {1 \over 2},y + {1 \over 2},\sqrt {1 - \left( {x + {1 \over 2}} \right)^2 - \left( {y + {1 \over 2}} \right)^2 } } \right)$$

and $V = \phi \left( U \right)$ which is the projection onto the upper hemisphere of the open disc of radius 1/4 centered at the point $\left( {{1 \over 2},{1 \over 2}} \right)$ in the x-y plane. Then $\phi $ parameterizes $Y$ around $x$ with $0 \in U$ and $\phi \left( 0 \right) = x$ . Furthermore, $ \tilde V = V \cap X $ is the open segment of $X$ that passes through $V$ . And since the canonical immersion (not "canonical inclusion") in our case is $j: {\rm I\!R}^1 \to {\rm I\!R}^2$, we have $\tilde U = j^{ - 1} \left( U \right) = \left( { - {1 \over 4},{1 \over 4}} \right)$ .

The suggested function definition is $\tilde \phi :\tilde U \to \tilde V$ , $\tilde \phi \left( x \right) = \phi \left( {x,0} \right)$ . We claim this function is not well-defined because $\tilde \phi $ does not take $\tilde U$ to $\tilde V$ . The image $ \tilde \phi \left( {\tilde U} \right) $ is the projection onto the upper hemisphere of the open line segment between the points $\left( {{1 \over 4},{1 \over 2}} \right)$ and $\left( {{3 \over 4},{1 \over 2}} \right)$ in the x-y plane. This makes $ \tilde \phi \left( {\tilde U} \right) $ a segment of the circle obtained by slicing the sphere with the plane $y = {1 \over 2}$ . Meanwhile $\tilde V$ is a segment of the circle obtained by slicing the sphere with the plane $z = {1 \over {\sqrt 2 }}$ . Therefore $ \tilde \phi \left( {\tilde U} \right) \not\subset \tilde V $. In fact, when you sketch it out, you'll see that $ \tilde \phi \left( {\tilde U} \right) $ and $\tilde V$ intersect only at $x$ .

The proof in the original post may have been philosophically misguided as it seems that it was trying to use the Local Immersion Theorem from Section 1.3 to prove the inclusion theorem presented in Section 1.2.

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I'm pretty sure there's a problem with your proof. The function $\tilde \phi$ defined as $\tilde \phi \left( {a_1 , \ldots ,a_k } \right) = \phi \left( {a_1 , \ldots ,a_k ,0, \ldots ,0} \right)$ doesn't necessarily take $\tilde U$ into $\tilde V$. I was able to construct a "counterexample" where $\tilde \phi \left( {\tilde U} \right) \not\subset \tilde V$. Unfortunately, I can't see how to modify your proof to fix it.

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  • $\begingroup$ I'd be glad to post the counterexample I cooked up. It involves a circle as a subspace of $S^2$. there might be simpler counterexamples. $\endgroup$ – TJCrow Jan 7 '17 at 23:50
  • $\begingroup$ please do post your counterexample. $\endgroup$ – Alex Ortiz Sep 22 '17 at 3:37
  • $\begingroup$ I posted the counterexample as another answer. It's showing just above this. $\endgroup$ – TJCrow Sep 26 '17 at 2:11

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