Homeomorphic Topological Spaces with the Subspace Topology

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.

• The only thing I can think that is missing is to actually apply the definition of "homeomorphic". – Lee Mosher Mar 4 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.

• 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? – callista Mar 3 at 23:28
• 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 $H$. – Lee Mosher Mar 3 at 23:42
• 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. – callista Mar 4 at 0:05
• I've added a bigger hint. – Lee Mosher Mar 4 at 0:52
• Great thank you! – callista Mar 4 at 1:55