For local diffeomorphisms,is the $f(U)$ open in range assumption redundant

differential-geometrygeneral-topologygeometrymanifoldssmooth-manifolds

My book is An Introduction to Manifolds by Loring W. Tu.

From Wikipedia: Local diffeomorphism:

For $X$ and $Y$ differentiable manifolds. A function $f:X\to Y$,
is a local diffeomorphism, if for each point x in X, there exists an open set $U$ containing $x$, such that $f(U)$ is open in Y and $f|_{U}:U\to f(U)$,
is a diffeomorphism.

(I guess the "$f|_U$" is not $f|_U:U \to Y$ but rather $f|_U$ with restricted range $\tilde\{f|_U\}: U \to f(U)$.)

For $X$ and $Y$ instead smooth, and not merely differentiable manifolds, is the assumption that $f(U)$ is open in $Y$ actually redundant?

  • I think we must have $f(U)$ at least a smooth (embedded or regular) submanifold of $Y$ since it wouldn't make sense for $\tilde\{f|_U\}$ to be a diffeomorphism otherwise. Also, I guess we must have $X$, $U$ and $f(U)$ to be of the same dimension.

  • I think that $f(U)$ is open in $Y$ follows from smooth invariance of domain given in Remark 22.5 on Theorem 22.3, which relies on Definition 22.1.

  • Then again I think smooth invariance of domain is not applicable because $X$ and $Y$ are not given as the same dimension. Either I'm missing something or Tu has a different definition of local diffeomorphism (defined in Section 6.7 and further described in Remark 8.12). For example, Wikipedia's local diffeomorphisms are open maps. I'm not sure Tu's local diffeomorphisms are too. Update: I think it's a different definition. See here.

  • I just noticed that Theorem 6.26 and Remark 8.12 have "same dimension". I think Tu's definition is the same as Wikipedia's assuming smooth manifolds of the same dimension.

Best Answer

No, it is not redundant. Any embedding is a diffeomorphism onto its image (so you could just take $U=X$ for every point in this case), which, in general, is not an open set in the codomain.

Now, if you assume $\dim X=\dim Y$, with $f(U)$ and embedded submanifold of $Y$, then, as you said, $\dim f(U)=\dim X$ and so you get $\dim f(U)=\dim Y$. You can see here that this does indeed imply that $f(U)$ is open in $Y$. So the assumptions would be equivalent but, honestly, I find the condition of $f(U)$ being open much more intuitive in this case.

I hope this helps (with Question $1$ at least).

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