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I have struggle with understanding details of the following construction: Let $M$ be a smooth manifold. consider $F:=C^{\infty}(M,\Bbb R)$, Now consider the following $f:M\to \Bbb R^F$!!!

What is going on? First we collect all real value and smooth maps to a set $F$, then we consider $\Bbb R^F$ that is $\{h|h:F\to \Bbb R\}$ then we consider $f:M\to \Bbb R^F$. What is happening here? What are these trying to explain?

Eric Wofsey
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C.F.G
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    Think of vector fields: they associate to every point a derivation, i.e. a linear functional on $C^\infty(M,\mathbb{R})$, giving an example of $f:M\to \mathbb{R}^F$ –  Nov 23 '20 at 14:02
  • So why in such a complicated way? $f:M\to TM$ isn't more intuitive? what is the advantage of the above construction? – C.F.G Nov 23 '20 at 15:26
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    That depends on what is the aim of the construction is: if you provide more context, I may be able to tell you more –  Nov 23 '20 at 15:37
  • This is from page 10 of Milnor's char. classes. There he want to show that $M$ is a submanifold of $\Bbb R^F$. and I wonder why $\Bbb R^F$? – C.F.G Nov 23 '20 at 17:02
  • @C.F.G The point is that we want a real vector space $V$ into which we can necessarily embed $M$. $\Bbb R^F$ is constructed in a complicated way, but the important takeaway in the context is that $\Bbb R^F$ is a vector space and $i:M \to \Bbb R^F$ is a diffeomorphism with its image. So, the question of existence is settled for any manifold $M$. – Ben Grossmann Nov 23 '20 at 17:31
  • @C.F.G If you are familiar with the isomorphism between a vector space and its double dual (or more generally, the embedding of a vector space in its double dual), then you might find it helpful to think of this embedding as being the natural generalization of this idea to the context of manifolds. – Ben Grossmann Nov 23 '20 at 17:36
  • @C.F.G If you have some exposure to category theory, you might also like to think of this as the of representation of an object given by the Yoneda lemma – Ben Grossmann Nov 23 '20 at 17:43
  • @C.F.G If the embedding of a vector space into its double dual is something you find unintuitive, you might find my answer here on the topic to be helpful. I believe that if you understand the embedding for vector spaces, then the embedding for manifolds will feel "natural". – Ben Grossmann Nov 23 '20 at 17:46
  • @BenGrossmann: After glancing to "Yoneda lemma" it seems that this construction is recurrent in mathematics, and I should take it easy although it is cumbersome a bit. – C.F.G Nov 23 '20 at 19:37
  • But from smooth manifold point of view it is hard to see for me. – C.F.G Nov 23 '20 at 19:52
  • @C.F.G In two sentences, the punchline (for manifolds and vector spaces) is this: in many situations, it is convenient to identify a point in $x$ with the associated "evaluation" $\alpha_x$, which is defined by $\alpha_x(f) = f(x)$. This identification will always be injective and "structure preserving" (i.e. linear or differentiable) – Ben Grossmann Nov 23 '20 at 20:24

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