1. Originally Posted by Bruno J.
In general, if you are given a subset of a topological space, which is itself referred to as a topological space, then it is assumed to have the subspace topology. Any subset of a topological space inherits the ambiant topology in a natural way. Let $\displaystyle \iota : S^2 \to \mathbb{R}^3$ be the inclusion map. The topology on $\displaystyle S^2$ is the topology generated by the open sets of the form $\displaystyle \iota^{-1}(U)$, where $\displaystyle U$ is open in $\displaystyle \mathbb{R}^3$.

Now it is easy to solve the problem. Show that any open cover of a set in $\displaystyle S^2$ lifts to an open cover of the same set, considered as a subset of $\displaystyle \mathbb{R}^3$. Conversely, any open cover of some subset of $\displaystyle S^2$ (considered as a subset of $\displaystyle \mathbb{R}^3$) can be considered as an open cover of the same set as a set of $\displaystyle S^2$.

By the way, the notation $\displaystyle S^n$ for the $\displaystyle n$-sphere is universal!
Lifting maps are way, way overkill for a simple problem like this, why introduce it?

The more simplistic idea is as follows...

Let $\displaystyle X$ be a topological space and $\displaystyle Y$ a subspace. Then, $\displaystyle Z\subseteq Y\subseteq X$ is a compact subspace of $\displaystyle Y$ if and only if it's a compact subspace of $\displaystyle X$

To see this first assume that $\displaystyle Z$ is a compact subspace of $\displaystyle Y$ and let $\displaystyle \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an open cover $\displaystyle Z$ with the subspace topology inherited from $\displaystyle X$. Then, $\displaystyle U_\alpha=V_\alpha\cap Z$ where $\displaystyle V_\alpha$ is open in $\displaystyle X$. But, note that $\displaystyle V_\alpha\cap Z=V_\alpha\cap Z\cap Y=\underbrace{\left(V_\alpha\cap Y\right)}_{\text{ope}\text{n in }Y}}\cap Z$ and so $\displaystyle \left\{V_\alpha\cap Y\cap Z\right\}_{\alpha\in\mathcal{A}}$ has a finite subcover (since it's an open cover of $\displaystyle Z$ as a subspace of $\displaystyle Y$) from where the conclusion follows.

The other direction is similar.

2. I'm not sure what you're saying. I'm not speaking of "lifting" in the sense of lifting a path from a manifold to some covering! I'm just saying that every open set $\displaystyle o$ of $\displaystyle S^2$ (as a topological space) can be obtained as the intersection of some open set $\displaystyle o'$ of $\displaystyle \mathbb{R}^3$ with $\displaystyle S_2$ (as a subset).

What you wrote is exactly what I was suggesting.

3. I'd also like to point out that the map $\displaystyle U_\alpha \mapsto V_\alpha$ which I described does constitute a lift in the sense of category theory.

4. Ok, i think i solved this, more or less, thanx all for the help and ideas

And let me say again, that it was a mistype in the header, it shouldnt be "open set" just "any set".

S^2 is compact if it is closed and bounded, and as S^2 contains all its 'peripheral points' (that is probably not the right word in english, but i hope it speaks for itself). Also there exist a 'ball'
containing S^2, so S^2 is compact, therefore K is compact in S^2 if and only if K is closed.. Since K ⊆ S^ 2 ⊆ R^3, then K will be compact in R^3 <=> K is closed.

5. Great.. I completely missed Drexel28's answer, which was more or less what i was after... Anyhoo i think its more or less solved now, thanx all

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