Let $\mathbb{R^3}$ be given the standard topology. Let $P$ be a sextic parabloid and $H$ be the circular half-hyperboloid in $\mathbb{R^3}$ defined by

$P = {(x,y,z) ∈ \mathbb{R^3} : y=x^6 + z^6}$

$H = {(x,y,z) ∈ \mathbb{R^3} : z^2 -1= x^2 +y^2, z \geqslant 0}$

Consider $P$ and $H$ as topological spaces with the subspace topology. Prove that $P$ and $H$ are homeomorphic.

I am struggling with this question and don't really know where to begin. I know how to prove that elementary functions are homeomorphic but have no idea how to do this for $P$ and $H$. Any help will be much appreciated.

So far, I have:

For the function, $F:\mathbb{R^2} \rightarrow P$ given by $F(x,z) = (x, x^6 + z^6, z)$ is a homeomorphism since it is continous and bijective, and its inverse, $F^{-1}:P \rightarrow \mathbb{R^2}$, given by $F^{-1}(x,x^6 + z^6,z)=(x,z)$ is also continous. I've done the same thing for the function G for the space H, but don't know where to go from here.

  • $\begingroup$ The only thing I can think that is missing is to actually apply the definition of "homeomorphic". $\endgroup$ – Lee Mosher Mar 4 '19 at 2:38

Let me answer with a hint.

The equation $y = x^6 + z^6$ is the graph of a function $y=f(x,z)$ which is clearly defined defined for all $(x,z) \in \mathbb R^2$.

Similarly, the equation $z = \sqrt{x^2 + y^2 + 1}$ is again the graph of a function $z=g(x,y)$, again defined for all $(x,y) \in \mathbb R^2$.

Visualizing those graphs, what familiar topological space are they both homeomorphic to?

For a bigger hint: Consider the functions $$F(x,z) = (x,f(x,z),z) $$ and $$G(x,y) = (x,y,g(x,y)) $$ Describe the domain and image of each. Prove that each is continuous and one-to-one. Then prove that each of their inverse functions, from their image to their domain, is continuous.

  • $\begingroup$ So should I construct a function that's homeomorphic to both P and H defined only on x as a sort of bridging function? And are they both homeomorphic to a parabola? $\endgroup$ – callista Mar 3 '19 at 23:28
  • $\begingroup$ You should construct a topological space that is homeomorphic to both. To be homeomorphic is an equivalence relation. So if you can prove that $P$ is homemorphic to some topological space $X$, and if you can prove that $H$ is homeomorphic to the same topological space $X$, then it follows that $P$ is homeomorphic to $H$. $\endgroup$ – Lee Mosher Mar 3 '19 at 23:42
  • $\begingroup$ Ah right, that all makes sense but I'm still stuck on imagining what this space X could possibly be. I've sketched P and H out and unfortunately it still hasn't really helped. $\endgroup$ – callista Mar 4 '19 at 0:05
  • $\begingroup$ I've added a bigger hint. $\endgroup$ – Lee Mosher Mar 4 '19 at 0:52
  • $\begingroup$ Great thank you! $\endgroup$ – callista Mar 4 '19 at 1:55

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