If $f$ is a continuous bijective map between two manifolds $M$ with dimension $m$ and $N$ with dimension $n$. Why this map does not necessarily preserve the dimension i.e $m\not = n$? What is a good example about that?
continuous bijective maps between manifolds
2
$\begingroup$
general-topology
differential-geometry
manifolds
-
3What exactly is your definition of "manifold"? Particularly, is a "manifold" second-countable? (If not, write an arbitrary manifold $M$ as the image of the $0$-manifold obtained from the discrete topology on the set $M$.) – 2017-01-08
-
0manifolds here are to be Hausdorff – 2017-01-08
-
2Okay, but *are they to be second countable?* – 2017-01-08
-
0no not necessarily – 2017-01-08
-
3Then Andrew's comment gives you examples, and unless I overlook something, all examples are built from that. – 2017-01-08
-
0so if $M$ and $N$ are second countable then $m$ must equal $n$? – 2017-01-08
-
3Yes. (Unless you count the silly example $M = \varnothing = N$, but considered as a manifold of different dimensions.) If $f \colon M \to N$ is continuous and injective, then $m \leqslant n$. If $m < n$, then $f(K)$ is a compact subset of $N$ with empty interior for every compact $K\subset M$. Since $M$ is $\sigma$-compact, $f(M)$ is then meagre in $N$. – 2017-01-08
2 Answers
1
$\newcommand{\Reals}{\mathbf{R}}$Let $M$ be an arbitrary $n$-manifold with $n \geq 1$, and let $M'$ be the same underlying set with the discrete topology (an uncountable $0$-manifold). The identity map (on sets) $M' \to M$ is a continuous bijection from a $0$-manifold to an $n$-manifold.
Similarly, the $(n + k)$-manifold $\Reals^{n + k}$ may be written as the bijective image of the $n$-manifold $$ \bigcup_{x \in \Reals^{k}} \Reals^{n} \times \{x\}, $$ with $\Reals^{n}$ viewed as the prototypical $n$-manifold and $\Reals^{k}$ treated as a discrete manifold.
For a more interesting example, write a $2$-torus as a disjoint union of uncountably many real lines immersed as translates of an irrational winding.
3
In the case $n=m$ : Any injective continous maps $\mathbb R^n \to \mathbb R^n$ is an homeomorphism on its image. This is the classical invariance of domain theorem.
Since an immersion is not an embedding in general, my argument I did wrote is not true for $n < m$, as George Elencwajg said in the comments.
-
0Thanks ! I will remember to not answer a question too quickly, without being sure this is correct and it does answer the question. – 2017-01-09
-
0Dear N.H., bravo for your New Year good resolution :-) – 2017-01-09
-
0Thanks again for your helpful and encouraging comments, dear Georges !! – 2017-01-09