About a condition for an immersion to be an embedding

Question

I was trying to work on immersions embeddings in the case of smooth manifolds, and one sentence I read in a book seemed to imply that an immersion $f:X \rightarrow Y$ (in the sense of smooth manifolds) is an embedding if and only if for every point $y$ in $Y$ there exist arbitrarily small open neighborhoods of $y$ in $Y$ such that $f^{-1}(U)$ is connected. Is that true?

Answer 1

Since $X$ (as a manifold) is locally compact, there are two things you need to verify in order to prove that an immersion $f: X\to Y$ is an embedding:

1. $f$ is injective.

2. If $x_i\in X$ is a sequence which diverges to infinity (in the 1-point compactification of $X$) then $f(x_i)$ cannot converge to some $y=f(x)$, $x\in X$.

I will verify Part 1 and will leave you to check Part 2. As in your question, I assume that every $y=f(x)\in f(X)$ has a basis of neighborhoods $\{U_\alpha\}_{\alpha\in A}$ such that $f^{-1}(U_\alpha)$ is connected for every $\alpha\in A$.

Suppose that $f$ is not 1-1: There exist distinct points $x_1, x_2\in X$ with $f(x_1)=f(x_2)=y$. Pick a compact subset $K\subset X$ whose interior contains $x_1, x_2$. Since $f$ is an immersion, it is locally injective. Thus, $f^{-1}(y)\cap K$ is finite, equal $\{x_1, x_2,...,x_k\}$.

Since $X$ is Hausdorff, there exist pairwise disjoint neighborhoods $V_1,...,V_k$ of $x_1,...,x_k$ and $U_\alpha$ such that $$ f^{-1}(U_\alpha)\cap int(K) \subset V_1\cup ...\cup V_k. $$ But each intersection $f^{-1}(U_\alpha)\cap V_i$ contains $x_i$, hence, is nonempty. Thus, $f^{-1}(U_\alpha)$ is disconnected. A contradiction.

Answer 2

So I think that I finally figuring it out. Thanks for the help, trying to prove the injectivity first helped me understanding the problem. Here it is: (the ball without the closure on it is the open ball)

Let $M$ and $N$, be two manifolds $C^p$ of dimensions respectively $n$ and $k$, and let $f$, be an immersion of class $C^p$ from $N$ to $M$. ($p \geq 1$ $n \geq k \geq 1$).

Let's assume that for each point of the image of $f$, there exists a basis of neighborhoods of this point in $M$ $(V_q)_{q \in \mathbb{N} }$ such that $\forall q \in \mathbb{N}$ $f^{-1}(V_q)$ is connected and $V_q$ open.

Lemma: Let $x \in N$ and $(x_r)_{r \in \mathbb{N} } \in N^{ \mathbb{N} }$ such that $f(x_r)$ has $f(x)$ as a limit in $M$. We then have that $x_r$ has $x$ as a limit in $N$.

Proof: We'll show it by contradiction: Let $(x_r)_{r \in \mathbb{N}}$ be sequence of points in $N$ such that the limit of $(f(x_r))_{r \in \mathbb{N}}$ is $f(x)$ in $M$ and that there exists $U$, an open set of $M$ such that $x \in U$ and that $\forall r \in \mathbb{N}$ $x_r \notin U$ (we can assume that by just taking a sequence not verifying the consequence of the lemma and then just keeping an infinity of terms not converging to $x$, thus by extraction we'll get a sequence of terms avoiding an open subset containing $x$ but such that the limit of their images by $f$ is $f(x)$).

Let $(V_q)_{q \in \mathbb{N}}$ be the basis of neighborhood of $f(x)$ that the assumption on $f$ gives us.

We note $\phi :W_1 \longrightarrow B(0,1)$ (with $W_1 \subset U$ ) and $\psi: W_2 \longrightarrow B(0,1)$ be charts such that $\phi (x)=0$ and $\psi (f(x))=0$.

$\psi \circ f \circ \phi ^{-1}$ is an immersion. Let $p$ be a linear projection from $\mathbb{R}^n$ onto the image of the differential of this immersion at the point $0$. Thus $p \circ \psi \circ f \circ \phi ^{-1}$ is $C^p$ and its differential at $0$ is an isomorphism $0$ de $\mathbb{R}^k$ sur $F$. By the inverse function theorem, $\exists \epsilon \in ]0;1[$ and $X \subset B(0,1)$ in $\mathbb{R}^k$ such that: $p \circ \psi \circ f \circ \phi ^{-1}$ is a $C^p$ diffeomorphism from $X$ on $B(0, \epsilon)$ (the ball in $F$). We note $\theta$, its restriction from $X$ to $B(0, \epsilon)$.

Let $q\in \mathbb{N}$ be such that $V_q \subset \psi ^{-1} (p^{-1}(B(0, \dfrac{ \epsilon}{2})))$. $f^{-1}(V_q)$ being a non empty open connected subset of a manifold, it is pathwise connected. Also, there exists $r_0 \in \mathbb{N}$ such that: $\forall r >r_0$ $f(x_r) \in V_q$. We deduce from it that there exists $\gamma$ a path from $x$ to $x_r$ in $f^{-1}(V_q)$ for $r$ being great enough.

Thus, $\gamma$ has its images out of $U$ for $t$ big enough because $x_r \notin U$. We note $I$, the interval in $[0;1]$ containing $0$ such that $\gamma (s)$ is in $X$ for $s \in I$ (that is the connected component of $\gamma^{-1}(U)$ containing $0$).

Let $\phi ^{-1}( \theta ^{-1}(B(0,\epsilon /2)))=A$.

We note $t$, the minimum of $\gamma^{-1}(A^c)$ (it exists because $1$ is in it, and $A$ is an open subset of $X$ thus of $N$, thus $\gamma^{-1}(A^c)$ is closed and non empty in $[0;1]$). $t \in I$ because $A \subset \phi ^{-1}( \theta ^{-1}( \overline{(B(0, \epsilon /2)})))$ which is compact because $\phi^{-1}$ and $\theta^{-1}$ are continous, and $N$ being Hausdorff, it is a closed subset of $N$ thus $\gamma(t)$ is in there by minimality of $t$, and $\gamma (t) \in X$, and $[0;t[ \subset I$. Thus $[0;t] \subset I$.

$t$ being in $I$, by continuity $\theta ( \phi ( \gamma (t))) \in \partial B(0, \epsilon /2)$ because $\phi ^{-1}( \theta ^{-1} ( \overline{((B(0,\epsilon /2))})))$ is a close subset of $N$ and it is not in the inverse image of the open ball.

Thus: $p \circ \psi \circ f( \gamma (t))$ is in $B(0, \epsilon /2)$ and in its complement. It is a contradiction.

Proposition: $f$ is injective and $g=f^{ |Im(f)}$ is a homeomorphism.

Proof: To show that it is a bijection, we chose the sequence $x_r=x'$ with $f(x)=f(x')$. We have, from it and the lemma, that $x_r$ tends to $x$. Thus $x'=x$. The continuity of $g^{-1}$ can be deduced from the lemma.

I hope there isn't a mistake in what I wrote...