question about a proof of “ bijective, continuous function between connected, locally euclidean spaces is a homeomorphism” （using invariant domain)）

Question

problem 1:(as in here)

Let $X$ and $Y$ be connected, locally Euclidean spaces of the same dimension. If $f:X \rightarrow Y$ is bijective and continuous, prove that $f$ is a homeomorphism.

source: Conlon's Differentiable Manifolds Exercise 1.1.(4)

problem 2:

Let $X$ and $Y$ be connected, locally Euclidean,second countable spaces of the same dimension. If $f:X \rightarrow Y$ is bijective and continuous, prove that $f$ is a homeomorphism.

source:problem set of my manifold course.

my question:

1.about a answer of problem 1

i understand the sketch: for arbitrary $x$ and $f(x)$,construct a continuous injective map from their charts(open subsets in Euclidean space). apply invariant of domain, then it "squeezes" $f$ to be homeomorphism. $$f:U_x\rightarrow f(U_x)$$ and $f(U_x)$ is open, since it's a preimage of some open set in chart of $y$.then $f$ is a bijective local homeomorphism, hence it's easy to prove it's global.

my question is :why we need connectness? i failed to find which part in the proof used connectness.

my answer of question 1:

connectness make "dimension of locally Eucildean space" well-defined. see Conlon corollary 1.1.8 so in fact we don't need it, it‘s just convenient with it.

2.about problem 2.

it seems there is a way to prove it without invariance of domain.but i don’t know how to use connectness and second countability property.

i know something in general topology:

• second countability is related to paracompactness but the space in question 2 isn't Hausdorf.

• second countability implies open cover has countable subcover.

but again how to use connectness?

Edit1:

in problem2 $X$ and $Y$ don't need to have the same dimension.

Answer 1

solution to Problem 2 by my TA.

Sketch:

Let $\dim X=n$,$\dim Y=m$, we'll prove $n=m$ then it becomes problem 1.

we need second countability to apply Baire category theorem:

Baire category theorem:locally compact Hausdorff space is Baire space. In particular countable union of nowhere dense set is nowhere dense.

and locally Euclidean space is locally Hausdorff: we do all our arguments locally in the open neighborhood homeomorphic to Euclidean space, hence in fact we do have the Hausdorff property.

Proof:

for any $p\in Y$, consider in the open neighborhood $A$ of $p$ such that $A$ is homeomorphic to $R^m$, then the restriction of $f$ on $f^{-1}(A)$ is a continuous bijection, and A is locally compact Hausdorff space. We'll prove this restriction of $f$ is homeomorphism hence $f$ is a local homeomorphism, combining the fact that f is bijective to its image, we have $f$ is a homeomorphism.

Now for any $y\in A$, let $U$ be the open neighborhood of $y$ that is homeomorphic to $R^m$, then there is a open neighborhood V of $x=f^{-1}(y)$ homeomorphic to $R^n$, $f|_V$ is a continuous bijection.

Case $n>m$:

Let $\iota$ be the canonical immersion from $R^m$ to $R^n$. then $\iota\circ f|_V:V \rightarrow R^n$ is an injective continuous map, By invariance of domain this is a homeomorphic map to its image, hence its image is an open set, yielding a contradiction.

Case $n<m$:

By second countablity, $U$ is a countable union of open precompact sets that are homeomorphic to open balls in Eucildean space， we denote these open sets as ${U_i}$.

$g=f|_{\overline{U_i}}$ is a continous bijective map from compact set to Hausdorff space is homeomorphism.(here we use the fact that $A$ is Hausdorff space)

Denote the interior of $f(\overline{U_i})$ as $M_i$, if $M_i\neq \emptyset$ , then $g^{-1}|_{M_i}$ is a homeomorphsim from open set of a higher-dimensional Euclidean space to a lower-dimensional Euclidean space, hence we can derive a contradiction as in case $n>m$.

Otherwise all $f(\overline{U_i})$ is nowhere dense, $A$ is the countable union of all $f(\overline{U_i})$, then A is a nowhere dense set itself since A is a Baire space, yielding a contradiction. q.e.d