A claim about smooth maps between smooth manifolds

[aalma]

TL;DR Summary

Trying to prove a claim using the definition of a smooth map.

Given the definition of a smooth map as follows:
A continuous map $f : X → Y$ is smooth if for any pair of charts $\phi : U →R^m, \psi:V →R^n$ with $U ⊂ X, V ⊂Y$, the map $\phi(U ∩f^{-1}(V)) → R^n$ given by the composition
$$\psi ◦ f ◦ \phi^{-1}$$ is smooth.

Claim: A map $f : X → Y$ between two smooth manifolds is smooth iff for any open $U ⊂ Y$ and any $φ ∈ C^∞(U)$ the composition $f^{-1}(U)→ U → R \in C^∞(f^{-1}(U))$.

My question is how to show this claim?

For one side, I can use the definition above:

a function $φ∈C^∞(U)$ on an open subset $U⊂Y$
is a smooth function from $U$ to $R$.Therefore, we can take the charts $\phi:f^{−1}(U)→R^m$
and $ψ:U→R^n$ to be the identity maps, and we have: $$ψ◦f◦{\phi}^{−1}:R^m→R^n$$
which is a smooth map between Euclidean spaces. Composing this with $φ$ gives a smooth function on $$f^{−1}(U):
φ◦(ψ◦f◦{\phi}^{−1}):f^{−1}(U)→R$$
and this is the map $φ∘f$
as we choose $ψ,\phi$ to be the identity maps.
Does this seem fine?

For the other direction, we suppose that for any open $U ⊂ Y$ and any $φ ∈ C^∞(U)$ the composition $f^{-1}(U)→ U → R \in C^∞(f^{-1}(U))$. And we want to see that $f$ is smooth globaly (the definiton above). How to show this with charts..?

 

[quasar987]

TL;DR Summary: Trying to prove a claim using the definition of a smooth map.

is a smooth function from U to R.Therefore, we can take the charts ϕ:f−1(U)→Rm
and ψ:U→Rn to be the identity maps, and we have: ◦◦ψ◦f◦ϕ−1:Rm→Rn
which is a smooth map between Euclidean spaces. Composing this with φ gives a smooth function on ◦◦◦f−1(U):φ◦(ψ◦f◦ϕ−1):f−1(U)→R
and this is the map

The fact that you're considering taking "$\phi$ and $\psi$ to be the identity" suggests to me that the manifolds you're thinking about are subsets of euclidean space. Is this right? If this is not right, then what you wrote doesn't make sense. If it's right, then what you wrote does makes sense (although the word "inclusion" instead of "identity" is more appropriate) but is wrong because an open subset U of a manifold is generally not open in the ambiant euclidean space: think of the surface of a sphere in 3d space. An open subset of the sphere will have empty interior in R³. So, inclusions can't be used as charts.

TL;DR Summary: Trying to prove a claim using the definition of a smooth map.

For the other direction, we suppose that for any open U⊂Y and any φ∈C∞(U) the composition f−1(U)→U→R∈C∞(f−1(U)). And we want to see that f is smooth globaly (the definiton above). How to show this with charts..?

What do you mean by "smooth globally"? Differentiability is a local property. Hint: what happens when you take as $\varphi$ the components of a local chart around $f(p)$ ?

 

[aalma]

Yeah I just meant to show that f is smooth on M.
In the second part, you mean taking $p \in V\subset X$ such that $f(p)\in U\subset Y$ then how it is useful assuming that for any open $U⊂Y$ and any $\phi∈C^{\infty}$ the composition $\phi\circ f$ is in $C^{\infty}(f^{-1}(U))$?
If in the first part manifolds are not subsets of euclidean space, then how I am supposed to consider it?

The fact that you're considering taking "$\phi$ and $\psi$ to be the identity" suggests to me that the manifolds you're thinking about are subsets of euclidean space. Is this right? If this is not right, then what you wrote doesn't make sense. If it's right, then what you wrote does makes sense (although the word "inclusion" instead of "identity" is more appropriate) but is wrong because an open subset U of a manifold is generally not open in the ambiant euclidean space: think of the surface of a sphere in 3d space. An open subset of the sphere will have empty interior in R³. So, inclusions can't be used as charts.
What do you mean by "smooth globally"? Differentiability is a local property. Hint: what happens when you take as $\varphi$ the components of a local chart around $f(p)$

 

[quasar987]

Yeah I just meant to show that f is smooth on M.
In the second part, you mean taking $p \in V\subset X$ such that $f(p)\in U\subset Y$ then how it is useful assuming that for any open $U⊂Y$ and any $\phi∈C^{\infty}$ the composition $\phi\circ f$ is in $C^{\infty}(f^{-1}(U))$?
If in the first part manifolds are not subsets of euclidean space, then how I am supposed to consider it?

Regarding my hint, all I'm saying is that differentiability is a local property, so assume we want to show f is differentiable at a point p of M. Then pick a local chart $(V,\psi)$ around f(p). Notice that the hypothesis implies that the compositions $\psi^j\circ f$ are smooth. Can you complete the argument from there?

 

[quasar987]

If in the first part manifolds are not subsets of euclidean space, then how I am supposed to consider it?

In general, a smooth manifold is not a subset of euclidean space. It is a topological space that's locally homeomorphic to euclidean space and such that, on overlapping charts, the transition functions are smooth.

Among the manifolds defined in this general way, there are some that occur as subspaces of euclidean space but, in general, when people speak of manifolds, they are not assuming this. So I was thrown off by your "take $\phi$ as the identity" idea.

 

[aalma]

I

Regarding my hint, all I'm saying is that differentiability is a local property, so assume we want to show f is differentiable at a point p of M. Then pick a local chart $(V,\psi)$ around f(p). Notice that the hypothesis implies that the compositions $\psi^j\circ f$ are smooth. Can you complete the argument from there?

I actually do not see how to translate it in the definition of smooth map, using the assumption in the claim..

 

[quasar987]

I

I actually do not see how to translate it in the definition of smooth map, using the assumption in the claim..

What does it mean to say that the functions $\phi^j\circ f$ are smooth ?

What does it mean to say that the function $f$ is smooth at $p\in M$ ?

 

[aalma]

What does it mean to say that the functions $\phi^j\circ f$ are smooth ?

What does it mean to say that the function $f$ is smooth at $p\in M$ ?

$f:M→N$ is smooth at $p \in M$ if for any pair of charts $\phi:V→R^m$ at $p$ and $\psi:U→R^n$ at $f(p)$ with $V⊂M,U⊂N$, the map $a(V ∩f^{−1}(U)) → R^n$ given by the composition $b ◦ f ◦ a^{−1}$ is smooth. It is sufficient to show that $f$ is locally smooth. Choose $V_p=V\cap f^{-1}(U)$ so it is a neighborhood of $p$ then $V_p\subset f^{-1}(U)$ so can i say that $f^{-1}(U)\to U\to R$ reduced on $V_p$ is smooth?
I really cannot figure out how to use this assumption!.

 

[martinbn]

For the first part, the only thing is that the open set $U$, which is the domain of the smooth function, need not be a chart. Same for its inverse image. So you need to take smaller open subsets around any point in $U$.

For the second part. The standad way is to consider charts and pull back the coordinate functions. Since they are smooth, the map will be a smooth map.

 

[quasar987]

$f:M→N$ is smooth at $p \in M$ if for any pair of charts $\phi:V→R^m$ at $p$ and $\psi:U→R^n$ at $f(p)$ with $V⊂M,U⊂N$, the map $a(V ∩f^{−1}(U)) → R^n$ given by the composition $b ◦ f ◦ a^{−1}$ is smooth. It is sufficient to show that $f$ is locally smooth. Choose $V_p=V\cap f^{-1}(U)$ so it is a neighborhood of $p$ then $V_p\subset f^{-1}(U)$ so can i say that $f^{-1}(U)\to U\to R$ reduced on $V_p$ is smooth?
I really cannot figure out how to use this assumption!.

That is indeed the question. Working from the hypothesis that the functions $\psi^1\circ f,\ldots, \psi^n\circ f$ are smooth on $V_p$, can you conclude that $f$ is smooth on $V_p$ ?

It is completely tautological. Meaning that just by unraveling the definitions we can see that these are logically equivalent statements.

You've already told me that, by definition, $f$ will be smooth on $V_p$ provided that $F:=\psi\circ f\circ \phi^{-1}$ is smooth. This is just a map from an open subset of R^m to R^n so we know what it means for F to be smooth. It means that all of its components $F^1\,..., F^n$ are smooth. What is $F^j$ ? It is $\psi^j\circ f\circ \phi^{-1}$. Is this smooth? By definition, it means that $\psi^j\circ f$ is smooth. Conveniently, that's our hypothesis.

One thing that you may have overlooked and which is possibly making this seem more complicated than it is it that you said

$f:M→N$ is smooth at $p \in M$ if for any pair of charts $\phi:V→R^m$ at $p$ and $\psi:U→R^n$ at $f(p)$ with $V⊂M,U⊂N$, the map $a(V ∩f^{−1}(U)) → R^n$ given by the composition $b ◦ f ◦ a^{−1}$ is smooth.

This is true, but you can also replace the words "for any pair of charts" by "for SOME pair of chart". That is, if f is smooth relative to some chart, then it is smooth relative to EVERY chart. Prove it if this is new to you. It comes down to the fact that the transition functions $\psi \circ \phi$ are smooth.

 

[aalma]

You are using $\psi$ as $φ$ in the main question?
If the functions $φ\circ f$ are smooth on $V_p$ then can one say that if we take $φ^{-1}$ is also smooth so $b\circ f|_{V_p}\circ a^{-1}=b\circ φ^{-1}\circ (φ\circ f|_{V_p})\circ a^{-1}$ which is smooth since $φ^{-1}$ is, and also $φ\circ f$ is smooth and $b, a^{-1}$ are homeomorphisms. Thus, $f$ is locally smooth and since $p$ is arbitrary we have that $f$ is smooth on $M$ (sorry for using differnt symbols each time). Does it make sense?

 

[quasar987]

In post #8 you introduced the symbols $a$ and $b$ but I just assumed you made a mistake and you meant $a=\phi$, $b=\psi$. But now you are seemingly purposefully making a distinction between $a$, $b$, $\phi$ and $\psi$. What is going on?! If you really thought of $a$ and $b$ as different from $\phi$ and $\psi$ in post #8 then the first sentence of that post makes no sense, does it?

 

[aalma]

In post #8 you introduced the symbols $a$ and $b$ but I just assumed you made a mistake and you meant $a=\phi$, $b=\psi$. But now you are seemingly purposefully making a distinction between $a$, $b$, $\phi$ and $\psi$. What is going on?! If you really thought of $a$ and $b$ as different from $\phi$ and $\psi$ in post #8 then the first sentence of that post makes no sense, does it?

Yeah, I meant that (I changed the symbol in order not to get confused with phi given in the question..).
And for the other side, I think, if we assume that $f:M\to N$ is smooth then by def, for the charts $a: f^{-1}(U)\to R^m$ and $b=φ:U\to R$ where $U\subset N$ is open (and so $f^{-1}(U)\subset M$ is open as $f$ is continuous) then by def $φ\circ f\circ a^{-1}$ is smooth so $φ\circ f|_{f^{-1}(U)}\circ a^{-1\circ a}$ is smooth

 

[quasar987]

This is an absolute notational disaster. If you want to continue the discussion we need to start from scratch, use good notation and stick with it. But honestly, I gave the complete solution in post #10. Have you tried to understand it?

 

[aalma]

>1. We say that two charts, $a: U → R^n$ and $b : V → R^m$ are compatible if the bijection
$b ◦a^{−1} : a(U ∩V) → b(U ∩V)$ is a diffeomorphism.
>2. A continuous map $f : M → N$ is smooth if for any pair of charts $: U →R^m, b:V →R^n$ with $U ⊂ M, V ⊂N$, the map
$a(U ∩f^{−1}(V)) → R^n$ given by the $b◦ f ◦ a^{−1}$ is smooth.

Now, I am tryiing to show that:

>A chart $a : U → R^n$ is compatible with an atlas on $M$ iff for each open $V ⊂ U$ it establishes a bijection between $C^∞(V)$ and $C^∞(a(V))$, i.e.
$f : a(V) → R$ is smooth iff $f ◦a$ belongs to $C^∞(V)$.

$(\Rightarrow)$ Assume $a$ is compatible with the atlas on $M$. Let $V$ be an open subset of $U$. Then for any chart $b : W \rightarrow \mathbb{R}^m$ in the atlas that intersects $V$, we have the map $b \circ a^{-1} : a(U \cap W) \rightarrow b(U \cap W)$ is a diffeomorphism. Let $f : a(V) \rightarrow \mathbb{R}$ be a smooth function. We need to show that $f \circ a : V \rightarrow \mathbb{R}$ is smooth. Let $b : W \rightarrow \mathbb{R}^m$ be a chart in the atlas that intersects $V$ so that $V \cap W$ is non-empty.Then $f \circ a = (f \circ b^{-1}) \circ (b \circ a^{-1})$ on $a(U \cap W)$. Since $f \circ b^{-1}$ is smooth because $f$ is smooth on $a(V \cap W)$ and $f \circ b^{-1}$ is its restriction to $b(V \cap W)$ and $b \circ a^{-1}$ is a diffeomorphism because $a$ and $b$ are compatible charts, it follows that $\circ a$ is smooth on $V$. Do you think this is okay?

Whay about the other direction, how should I consider it?

 

[quasar987]

You have the right idea for "==>" but again, sorry to say but your post is a mess.

1) Very first line, a and b are chart on the same manifold so they should both map into R^m.

2) In ">2." you give the definition of a smooth map between manifolds. Yet this question is purely about functions from a manifold into R. This is causing confusion especially since b in this definition is a chart on N while everywhere else in your post, b is a chart on M. Just give the defintion of a smooth map M-->R.

3) In what you are trying to show, we don't need V at all. This is just overly complicating things. See (4) below.

4) You have the right idea for the solution but there is a mistake in the key identity. You wrote $f\circ a=(f\circ b^{-1})\circ (b\circ a^{-1})$. However, the RHS of this is $f\circ a^{-1}$, not $f\circ a$. But you have the right idea in that you'd like to use the hypothesis that $b\circ a^{-1}$ is smooth. So, what I suggest is this. Instead of trying to show what you set out to show, show this:

"A homeomorphism $a:U\rightarrow \mathbb{R}^m$ is compatible with the atlas on $M$ iff it establishes a bijection between $C^{\infty}(U)$ and $C^{\infty}(a(U))$ in the sense that a function $f:U\rightarrow \mathbb{R}$ is smooth iff $f\circ a^{-1}:a(U)\rightarrow \mathbb{R}$ is smooth."

Notice the difference from your formulation where for you, $f$ is a map on $a(U)$, whereas for me, it is a map on $U$. Logically, what you wrote is equivalent to what I wrote but my formulation makes the solution much more natural I think. Just apply your idea of jamming $b^{-1}\circ b$ in the middle of $f\circ a^{-1}$ and you will be able to apply the hypothesis that $b\circ a^{-1}$ is smooth as you'd envisioned.

 

[Shirish]

Even my understanding of differential geometry is primitive so it'll help to participate in this thread. But isn't a function $f$ defined to be smooth (at $p\in U$) to be such that $f\circ a^{-1}$ is smooth (at $a(p)$) for any chart $(U,a)$ containing $p$? Won't that automatically establish a bijection between $C^{\infty}(U)$ and $C^{\infty}(a(U))$?

I can say, let $\alpha:C^{\infty}(U)\to C^{\infty}(a(U))$ be such that $\alpha(f)=f\circ a^{-1}$. If $g\in C^{\infty}(a(U))$, then $g\circ a\in C^{\infty}(U)$ by definition since $(g\circ a)\circ a^{-1}$ is smooth on $a(U)$. This proves surjectivity and it's also easy to prove injectivity. I didn't need to use the compatibility condition anywhere, so I'm probably missing something.

Maybe I'm making some implicit assumptions. I guess this bijection trivially holds for any chart in a complete atlas?

 

[quasar987]

Even my understanding of differential geometry is primitive so it'll help to participate in this thread. But isn't a function $f$ defined to be smooth (at $p\in U$) to be such that $f\circ a^{-1}$ is smooth (at $a(p)$) for any chart $(U,a)$ containing $p$? Won't that automatically establish a bijection between $C^{\infty}(U)$ and $C^{\infty}(a(U))$?

I can say, let $\alpha:C^{\infty}(U)\to C^{\infty}(a(U))$ be such that $\alpha(f)=f\circ a^{-1}$. If $g\in C^{\infty}(a(U))$, then $g\circ a\in C^{\infty}(U)$ by definition since $(g\circ a)\circ a^{-1}$ is smooth on $a(U)$. This proves surjectivity and it's also easy to prove injectivity. I didn't need to use the compatibility condition anywhere, so I'm probably missing something.

Maybe I'm making some implicit assumptions. I guess this bijection trivially holds for any chart in a complete atlas?

I think you basically answered yourself there. There is a manifold, and an atlas on it. So that's a collection of local homeomorphisms into euclidean space such that their composition on overlaps are diffeomorphisms. Such a system allows one to define smoothness of a function $f:M\rightarrow\mathbb{R}$ at a point $p\in M$. Namely, it is "smooth at $p$" if $b^{-1}\circ f$ is smooth for one, hence for all, chart (b,V) of the atlas around p. The "hence for all" is due to the compatibility condition that the transition maps $b\circ a^{-1}$ are diffeomorphisms.

So now the question our friend is trying to answer says this: "Pick a homeomorphism $a:U\rightarrow a(U)\subset \mathbb{R}^m$. Is (a,U) compatible with the atlas on M? That is, are all the transition maps $b\circ a^{-1}$ diffeomorphisms when $b$ comes from the atlas? Show that the answer is yes if and only if a function $f:U\rightarrow \mathbb{R}$ is smooth (relative to the atlas!) iff $f\circ a^{-1}$ is smooth (in the ususal sense for functions on R^m)

 

[aalma]

Thanks! I think I got that direction. It is just that I can say: $f\circ a^{-1}$ is smooth iff $f\circ b^{-1}$ is (since $b\circ a^{-1}$ is just a homoeomorphism) which is smooth iff $f$ is by the def of smooth function $f:U\to R$.
Can you explain the idea in the other direction?

 

[quasar987]

"A homeomorphism a:U→Rm is compatible with the atlas on M iff it establishes a bijection between C∞(U) and C∞(a(U)) in the sense that a function f:U→R is smooth iff f∘a−1:a(U)→R is smooth."

I think we might need to add the condition that a is smooth here.

For $\Leftarrow$ we assume that $a$ has the property that a function $f:U\rightarrow \mathbb{R}$ is smooth iff the composition $f\circ a^{-1}$ is smooth. From this hypothesis, we want to show that $b\circ a^{-1}$ and $a\circ b^{-1}$ are smooth for any chart $b$ in the atlas. As I explained in post #10 of the other thread, the main "ingredient" is the observation that the maps $b\circ a^{-1}$ and $a\circ b^{-1}$ are smooth iff their components are smooth. What is the $j$th component of $b\circ a^{-1}$? It is just $b^j\circ a^{-1}$. Ok well this is of the form $f\circ a^{-1}$ for the function $f=b^j$. This is how we connect with the hypothesis.

 

[mathwonk]

assume X and Y are euclidean space. then the only if part of the claim follows from the chain rule, and the if part follows by the definition of smoothness, i.e. the hypothesis implies all the component functions are smooth, so the map is smooth. Then charts allow this statement to be transferred back to the manifold in a trivial but confusingly complicated way, always using just the chain rule.

so my advice is, prove it first for X and Y equal to Euclidean space. Then see if you can't prove it in general.