Universal property of submanifold

Question

Updated: Apologize. I realize that this question is not worth answering.

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It's known that the subspace of a topological space shares the universal property as follows: let $X, Y$ be topological spaces and $i\colon X\hookrightarrow Y$ be an injective continuous mapping. Thus, $X$ is homeomorphic to the topological subspace $i(X)$ of $Y$, iff the following property is satisfied: for every topological space $Z$ and every set mapping $f\colon Z\rightarrow X$, $i\circ f$ is continuous iff $f$ is continuous, with some descriptions here.

Thus, I'm wondering whether there is an analogue for submanifold. This book told us(pp 11-12), this universal property cannot be transplanted to the category of manifold. That's to say, let $N$ and $M$ be smooth manifolds, with an injective smooth mapping $i\colon N\hookrightarrow M$. Even if the following property is satisfied.

Property 2.9.1. For any manifold $Z$ a mapping $f\colon Z\rightarrow M$ is smooth if and only if $i\circ f\colon Z\rightarrow N$ is smooth.

$N$ may not be diffeomorphic to the embedded submanifold $i(N)$ of $M$. And they gived the following counterexample.

Remark 2.13. If $f\colon \mathbb R\rightarrow \mathbb R$ is a function such that $f^p$ and $f^q$ are smooth for some $p, q$ which are relatively prime in $\mathbb N$, then $f$ itself turns out to be smooth, see [Joris, 82]. So the mapping $i\colon t\mapsto (t^p, t^q), \mathbb R\rightarrow \mathbb R^2$, has the property 2.9.1, but $i$ is not an immersion at $0$.

However I cannot see why remark 2.13 is an counterexample: since if we take $Z := \{(x^p, x^q): x\in \mathbb R\}$ as the embedded submanifold of $\mathbb R^2$. The embedded inclusion $j\colon Z \rightarrow \mathbb R^2$ is smooth. However $f\colon Z \rightarrow \mathbb R\colon (x^p, x^q)\mapsto x$ is in general not smooth (e.g. $f\colon (x^3, x^2) \mapsto x$ isn't smooth), and $i\circ f = j$ is smooth.

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Clearly, I made a mistake. Here I would like to summarize

Therefore, my question is:

1. Is there something wrong with my understanding?
2. Moreover, I believe the proof for the universal property of topological subspace is still work for submanifold. But why I cannot find anything about this characterization in other books about manifold?

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3. Is there a characterization of embedded submanifold using $C^\infty(M)$? For example, using $i^*\colon C^\infty(N)\rightarrow C^\infty(M)$?

Answer 1

Consider a smooth injective map $i\colon \mathbb R\to \mathbb R^2$ given by $i(t)=(t^3, t^2)$. The image $i(\mathbb R)$ is the affine variety $\{(x, y)\in \mathbb R^2 \mid x^2-y^3=0 \}$, where $(0, 0)$ is a singular point.

However, $i$ has Property 2.9.1: for every mapping $f\colon Z\to \mathbb R$ the composition $i\circ f\colon Z\to \mathbb R^2$ is smooth if and only if the mapping $f$ is smooth.

$\Leftarrow$ If $f$ is smooth, we see that $i\circ f=(f^3, f^2)$ is smooth.

$\Rightarrow$ If $i\circ f = (f^3, f^2)$ is smooth, then both $f^3$ and $f^2$ are smooth. Hence, according to Joris thorem, cited in Remark 2.13, $f$ is also smooth.

Regarding Question 3, I do not have an answer. The closest I have seen to this question is the discussion Diffeological submanifolds and their friends by Yael Karshon, David Miyamoto and Jordan Watts.

Answer 2

Long comment but not a complete answer (it's been a long time since I looked at this carefully and unfortunately I can't remember any reference ):

Regarding 3: First, $\iota^*$ maps $C^\infty(M)$ to $C^\infty(N)$ by restriction. You have it backwards.

Second, the idea is to look at the kernel of that map. $N$ is a codimension $k$ submanifold of $M$ if and only if for each $p \in N$, there exist smooth functions $f^1, \dots, f^k$ such that $$N = \{ f^1 = \cdots = f^k = 0\}$$ on a neighborhood of $p$ in $M$ and $\{df^1(p), \dots, df^k(p)\}$ is a linearly independent set in $T^*_pM$. I believe this can be expressed in terms of the ideal $$\mathcal{I} = \{ f \in C^\infty(M)\ :\ \left.f\right|_N = 0 \} $$ without any explicit reference to the differentials.

Another observation is that if $$ N = \{ f_1 = 0 \} = \{ f_2 = 0 \} $$ in a neighborhood of $p$ and $df_1(p)$ and $df_2(p)$ are both nonzero, then there exists a smooth function $g$ such that $g(p) \ne 0$ and $f_2 = gf_1$ in a neighborhood of $p$.

Based on the table on page 12 of these notes, it looks like the characterization is that $\mathcal{I}$ is a prime ideal. You might want to try to prove that.