Let $M,N$ be manifolds with $\dim M = \dim N$. If $f:M\to N$ is an immersion then $f$ is open.

I thought that I have solved it, but then I thought there could be a mistake:

Let $p\in M$. As $f$ is an immersion, $df_p$ is injective. Hence it is an isomorphism because $\dim M= \dim N$. By the Inverse Function Theorem (for manifolds), $f$ is a local diffeomorphism at $p$.

Let $A\subseteq M$ be an open set. For every $p\in A$, let $U_p\subseteq A$ be an open set such that $f\restriction_{U_p}:U_p\to f(U_p)$ is a diffeomorphism. I thought that as $f\restriction_{U_p}$ is a diffeomorphism, it is a homemorphism, hence $f(U_p)$ is open and $f(A)$ is union of open sets.

But then I remembered that $f(U_p)$ is only open in $f(U_p)$... which we already knew. I mean, $f\restriction_{U_p}$ is open as a function $U_p\to f(U_p)$, where $f(U_p)$ has the subspace topology, so that doesn't mean $f(U_p)$ is open in $N$. Right?


2 Answers 2


It suffices to show that $f$ maps "small" enough open sets to open sets. For that, given $p \in M$, let $U$ be a chart around $p$ such that $f$ is locally the identity. That is, $U$ and $V$ are such that the following diagram commutes.

$\require{AMScd}$ \begin{CD} U @>f>> V\\ @V \phi V V @VV \psi V\\ U'\subset \mathbb{R}^n @>>Id> V' \subset \mathbb{R}^n. \end{CD}

This is a consequence of the local form of immersions (since $\dim M=\dim N$! Otherwise, $Id$ should be replaced by a inclusion). Now, $f(U)=\psi^{-1} \circ Id \circ \phi$, and each one is an open map ($\phi$ and $\psi^{-1}$ by assumption about charts, and $Id$ since it is $Id$).


Immersions of manifolds, with dimensions, with equal dimensions are open maps:

  1. Local homeomorphisms are open maps.

  2. Local diffeomorphisms, since they are local homeomorphisms, are open maps.

  3. Submersions are open maps.

  4. For manifolds $M$ and $N$ that have dimensions and for $\dim M = \dim N$, we have by this that the following are equivalent for each $p \in M$ and for any smooth map $f: M \to N$

Local diffeomorphism at $p$, Immersion at $p$, Submersion at $p$.

  1. By (4), for manifolds $M$ and $N$ that have dimensions and for $\dim M = \dim N$, we have that the following are equivalent for any smooth map $f: M \to N$

Local diffeomorphism, Immersion, Submersion.

  1. Therefore, by (5) and (2), or by (5) and (3), immersions are open maps if the domain and range have equal dimension.

P.S. I think this converse is true: an open immersion implies dimensions are equal, so open immersions are equivalent to local diffeomorphisms. I ask about this here. (Note: When I say "converse", I don't refer to "open smooth maps of the same dimension are immersions".)


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