Vtyjkyuk

Can we find a homeomorphism from the square $Q_2$ of side length $2$ centered on the origin and the unit circle $S^1$?

We can easily define a map $r:Q longrightarrow S^1$ by

$$(x,y) mapsto bigg(fracxsqrtx^2+y^2,fracysqrtx^2+y^2 bigg)$$ which is a radial projection onto the unit circle, but how can we define $r^-1$ and show that it is continuous?

Thoughts

I think we may define the inverse map as

$$(x,y) mapsto bigg(fracxsqrt2max│x│, │y│ , fracysqrt2max │x│, │y│bigg)$$At least intuitively this maps to a square, and the $frac1sqrt2$ term scales appropriately. However, how can we demonstrate these are continuous maps, thus demonstrating that $Q_2$ and $S^1$ are homeomorphic?

Any help would be appreciated. Regards, MM.

Let me outline a proof; I shall leave the details of this proof as exercises.

Exercise 1: Prove that the unit circle $S^1$ is homeomorphic to the space obtained from the unit interval $[0,1]$ by identifying the endpoints $0$ and $1$. (Hint: the exponential mapping $thetato e^itheta$ can be composed with a linear mapping to give a homeomorphism.)

Let $C$ be the image of a simple closed curve in the plane, that is, let $C$ be the image of a continuous function $f:[0,1]to mathbbR^2$ such that $f:(0,1)tomathbbR^2$ is injective and $f(0)=f(1)$.

Exercise 2: Prove that $C$ is homeomorphic to the unit circle $S^1$. (Hint: show that $f:[0,1]to C$ induces a continuous bijection $tildef:S^1to C$ using Exercise 1 and then show that $f$ is an open mapping using the compactness of $S^1$ and the fact that the plane is Hausdorff.)

We now need an elementary lemma of point-set topology:

Exercise 3: Let $X$ be a topological space and let $X=Acup B$ where $A$ and $B$ are closed subspaces of $X$. Prove that if $g:Ato Y$ and $h:Bto Y$ are continuous functions into a topological space $Y$ such that $g(x)=h(x)$ for all $xin Acap B$, then there is a unique continuous function $f:Xto Y$ such that $f(a)=g(a)$ and $f(b)=h(b)$ for all $ain A$ and $bin B$.

Finally, we can prove the result of your question:

Exercise 4: Prove that the unit square is the image of a simple closed curve in the plane and conclude that it is homeomorphic to the unit circle. (Hint: you can use Exercise 3 to "glue" continuous mappings together.)

Exercise 5: Give examples of images of simple closed curves in the plane (using Exercise 3) and conclude (using Exercise 2) that they are homeomorphic to the unit circle $S^1$.

I hope this helps!

I'm not sure why you need algebraic formulas. Define each map $f$ to be radial projection onto the appropriate curve (from the origin) and find continuity by the property "$forall$ neighborhood $U$ of $f(x)$ $exists$ a neighborhood of $x$ whose image is contained in $U$", which is easy to check by eye.

It seems off to me. For example, in your second map, the point $(1,0)$, which is on the unit circle, gets mapped to $(frac1sqrt2,0)$, which is not on the square.

If I were you, I would write out the four different maps for each side of the square - I think it simplifies the process, especially since once you remove the max terms, the function is easier to work with.

If you choose to go about it in a computational manner, you can prove that both of the maps are continuous by noting that they are compositions of continuous functions.

Following in line with @Amitesh Datta's answer above, rather than proving the homeomorphism directly there are simpler ways that one can arrive at the desired result.

This following technique uses a fairly high-powered result (in the sense that once you know it, you can prove a lot of things are homeomorphic quite easily), but I think it's a very useful result that anyone doing topology should be aware of.

Definition: An open $n$-cell is any topological space that is homeomorphic to the open unit ball $mathbbB^n$. A closed $n$-cell is any topological space homeomorphic to $overlinemathbbB^n$

This is the high-powered result that I mentioned earlier.

Theorem: If $D subseteq mathbbR^n$ is a compact convex subset with nonempty interior, then $D$ is a closed $n$-cell and it's interior is an open $n$-cell. In fact given any point $p in operatornameInt(D)$, there exists a homeomorphism $F: overlinemathbbB^n to D$ that sends $0$ to $p$, $mathbbB^n$ to $operatornameInt(D)$ and $mathbbS^n-1$ to $operatornameBd D$.

Now note that if we let $A subseteq mathbbR^2$ be the (solid) square of side length $2$, then $A$ is easily seen to be a convex subset, and is thus a closed $2$-cell by the theorem above. Moreover the square $Q_2$ is the topological boundary of the solid square $A$ and the theorem above then immediately shows that $operatornameBd(A) = Q_2 cong S^1$ as desired.