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Following this question: Sphere homeomorphic to plane?

I understand that a sphere is not homeomorphic to the plane because the sphere is compact and the plane is not. But why is the sphere not homeomorphic to $[0,1]^2$?

This question interests me since if these two were homeomorphic, map projections would not be necessary. All previous mathematical answers I've seen to this question rely on the non-compactness of the plane, whereas the paper and screens maps are actually projected on in real-life are in fact compact.

Connor Brown
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    are you allowed to use algebraic topology tools (fundamental group for example)? – Sine of the Time Oct 13 '23 at 15:27
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    The sphere and $[0, 1]^2$ are not homeomorphic for more "algebraic topological reasons". One reason is that if you remove a point from the middle of $[0, 1]^2$, then it it no longer simply connected, but if you remove any point from the sphere it remains simply connected. However, more importantly, homeomorphism isn't the obstacle with map projections! For example, the closed hemisphere is homeomorphic to $[0, 1]^2$, but you still can't unwrap it onto a flat map. Homeomorphism allows too much "bending"! The concept you need is "local isometry" from geometry. – Izaak van Dongen Oct 13 '23 at 15:31
  • For that matter, the open square isn't homeomorphic to the closed square, but obviously you wouldn't have any trouble doing a map projection of one onto the other (not including the boundary, of course). – Brian Tung Oct 13 '23 at 15:51
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    Thank you, Izaak van Dongen, this feels exactly like the answer I was looking for! For completeness (and for others who come across my question later), it also may be worth looking at this thread that explains why there is no local isometry between points on a sphere and points on a plane: https://math.stackexchange.com/questions/748872/local-isometry-of-sphere – Connor Brown Oct 13 '23 at 15:59

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Here are some ways to see that $S^2$ and $[0,1]^2$ aren't homeomorphic:

  • $[0,1]^2$ is contractible while $S^2$ isn't. Indeed a contraction of $[0,1]^2$ is easy to write down, while $S^2$ is not contractible since it has a non-trivial second homology group.

  • If there would exist a homeomorphism $\phi : [0,1]^2 \to S^2$ and if $x$ is an interior point of $[0,1]^2$ then $\phi$ would still be a homeomorphism $[0,1]^2 \setminus \{ x \} \to S^2 \setminus \{ \phi(x) \}$, where both spaces have the subspace topology. But now $[0,1]^2 \setminus \{ x \}$ has a nontrivial fundamental group, while the fundamental group of $S^2 \setminus \{ \phi(x) \}$ is still trivial.

  • Take a look at this question if you want a proof which doesn't use algebraic topology (by showing that $S^2$ is not contractible without homology or cohomology).

As for your other point (if I understand it correctly), note that maps of the globe do not give homeomorphisms $[0,1]^2 \to S^2$ since on world maps there are no points corresponding to the north pole or the south pole. (That's one issue - another is that the inverse of a world map is not continuous.)

Steven
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$S^n$ is homogeneous, that is for $x, y\in S^n$ there is a homeomorphism $h:S^n\to S^n$ with $h(x) = y$. Indeed, from linear algebra we know that for each $x, y\in S^n$ there is an isometry $A$ sending $x$ to $y$ (for any two orthonormal bases $e_1, ..., e_n$ and $f_1, ..., f_n$ there is an isometry $A$ with $Ae_i = f_i$. Simply take $e_1 = x, f_1 = y$ and extend to orthonormal bases).

But $[0, 1]^n$ is not homogeneous, since points of the "interior" $(0, 1)^n$ and the "boundary" $[0, 1]^n\setminus (0, 1)^n$ can't be mapped to each other by a self-homeomorphism (here $n\geq 1$). But curiously, the Hilbert cube $[0, 1]^\omega$ is homogeneous.

Another reason why the two aren't homeomorphic is that $[0, 1]^n$ has the fixed point property, but $S^n$ doesn't, since the map $x\mapsto -x$ doesn't have a fixed point.

Jakobian
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    It's easy to show $[0,1]^n$ isn't homogeneous using algebraic topology. I can see one argument for this that uses elementary topology but it may be good to add a sketch of such an argument (a similar remark applies to showing $[0,1]^n$ has the fixed point property). – FShrike Oct 13 '23 at 16:25
  • @FShrike No sketch needed. The argument containts all relevant ingredients -- It is clear that a finite dimensional manifold with non-empty boundary cannot be homogoneous. – Moe Khaled Bin-Lateef Oct 14 '23 at 23:38
  • @MoeKhaledBin-Lateef Yes, it's clear to me because I know algebraic topology. My original comment was slightly in error, my idea for the elementary sketch doesn't really work and in fact relies on algebraic topology yet again. So please tell me, how is it clear (using methods of elementary topology)? – FShrike Oct 14 '23 at 23:42
  • @FShrike Seems to me that boundary points and interior points do not intersect is elementary enough. – Moe Khaled Bin-Lateef Oct 15 '23 at 00:11
  • @MoeKhaledBin-Lateef It's intuitive, sure. But please, provide an elementary proof. I'm genuinely curious to see if there is one. As far as I'm concerned, it is easy to show... if you can show that a once-punctured open ball is not contractible. Which is easy to show... if you know algebraic topology. – FShrike Oct 15 '23 at 00:17
  • @FShrike Good question. I'll have to think about it. – Moe Khaled Bin-Lateef Oct 15 '23 at 00:19
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No, $[0,1]^2$ is contractible and $\mathbb{S}^2$ isn't.

Moe Khaled Bin-Lateef
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If they were homeomorphic, say by a homeomorphism by $f:S^2\rightarrow [0,1]^2$, then also are homeommorphic $S^2\setminus\{N\}$ and $[0,1]^2\setminus \{f(N)\}$. However, we know that $S^1\setminus\{N\}$ is homeomorphic to $\mathbb{R}^2$, so it is simply connected, but $[0,1]^2\setminus\{N\}$ is not simply connected, then they can not be homeomorphic.

Ygor Arthur
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    $[0,1]\times [0,1]$ minus a point can be simply connected if you remove a point from the boundary – Sine of the Time Oct 13 '23 at 15:35
  • So, perhaps we can get another proof like this: there are $P,Q \in [0,1]^2$ so that $[0,1]^2 \setminus P$ and $[0,1]^2 \setminus Q$ are not homeomorphic; but for any $P,Q \in S^1$, we have $S^1 \setminus P$ and $S^1 \setminus Q$ are homeomorphic. The idea is to avoid more advanced comcepts like "simply connected". – GEdgar Oct 13 '23 at 16:15
  • @SineoftheTime you can just pick another point, that is not an issue. Take any point outside the boundary and it is the image of some point in $S^1$. – Ygor Arthur Oct 13 '23 at 19:41
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    @YgorArthur I know, but that's not what is written in your answer – Sine of the Time Oct 14 '23 at 05:19
  • @SineoftheTime I see the confusion. For me, $N$ is any point, but I agree that it could be interpreted as the north pole with this notation. – Ygor Arthur Oct 16 '23 at 15:54