# Show $T$ is a homeomorphism from $E$ onto $T(E)$

Let $$E$$ be the open interval $$(0, 1)$$ with the usual metric topology and $$\Bbb R$$ be given the usual topology.

Let $$\Bbb R^J$$ be given the topology of pointwise convergence $$T_p$$, where the index set $$J$$ comprises all continuous functions $$f : E \to \Bbb R$$.

Define a function $$T : E \to \Bbb R^J$$ by $$T(x) = (f(x))_{f∈J}$$ . Suppose that the image set $$T(E)$$ is given the subspace topology as a subspace of $$(\Bbb R^J, T_p)$$.

I want to show that $$T$$ is a homeomorphism from $$E$$ onto $$T(E)$$ but am not sure what method to employ even hmm.

Is there an explicit map to find? or find continuous maps $$f$$ and $$g$$ s.t. $$f\circ g=g\circ f= i$$ where $$i$$ is the identity map?

First notice that $$T : E \to T(E)$$ is continuous. Indeed, let $$(x_\lambda)_{\lambda\in\Lambda}$$ be a net in $$E$$ converging to $$x \in E$$. Then also $$f(x_\lambda) \to f(x)$$ for every $$f \in J$$ so $$T(x_\lambda) = (f(x_\lambda))_{f \in J} \to (f(x))_{f \in J} = T(x)$$ in the product topology on $$\mathbb{R}^J$$.

We can construct the inverse for $$T$$. The inverse $$T^{-1} : T(E) \to E$$ is given by the canonical projection $$\pi_g : \mathbb{R}^J \to \mathbb{R}$$ (with appropriate domain and codomain) where $$g \in J$$ is the inclusion $$g(x) = x, \forall x \in E$$.

Indeed, for any $$x \in E$$ we have $$T^{-1}(T(x)) = T^{-1}((f(x))_{f \in J}) = \pi_g(f(x))_{f \in J} = g(x) = x$$ and for any $$T(x) \in T(E)$$ with $$x \in E$$ we then have $$T(T^{-1}(T(x))) = Tx$$ which shows that $$T^{-1}$$ is indeed the inverse for $$T$$. It is clearly continuous so we conclude that $$T$$ is a homeomorphism.

• Ah thank you! By constructing the inverse for T like that, does it suffice to prove T is an open map though? – Homaniac Mar 7 at 0:19
• @Homaniac $T$ is a homeomorphism if both $T$ and $T^{-1}$ are continuous. Equivalently, $T$ must be continuous and an open map. – mechanodroid Mar 7 at 0:21
• Ah thanks makes sense, I am needing further details/comments too for a suggested solution in another homeomorphism question,if you like to help :) math.stackexchange.com/questions/3129188/… – Homaniac Mar 7 at 0:28
• @Homaniac The existing answer already provides a pretty explicit homeomorphism. Do you have problems with showing it is a homeomorphism, or would you like a different map (not that one comes to mind now, I'll have another look tomorrow)? – mechanodroid Mar 7 at 0:33
• More of the showing that is a homeomorphism and how it matches up with the equivalence relation conditions given in that qns though I guess this only goes under the comments section over there, not really a full answer but thanks for the help~ – Homaniac Mar 7 at 0:38

You have the map $$T$$. This is the homeomorphism between $$E$$ and and $$T[E]$$.

So you need to show that $$T$$ is 1-1/injective. This comes down to the fact that if $$x \neq y$$ in $$E$$ there is a continuous function $$f: E \to \mathbb{R}$$ (so $$f \in J$$) such that $$f(x) \neq f(y)$$.

Then $$T$$ is shown to be a bijection between $$E$$ and $$T[E]$$.

$$T$$ is continuous because $$\pi_f \circ T = f$$ is continuous for every $$f \in J$$.

$$T$$ has a continuous inverse too. This is the trickest part. This turns out to be equivalent to the fact that if $$C \subseteq E$$ is closed and $$x \notin C$$, there is a function $$f \in J$$ such that $$f(x) \notin \overline{f[C]}$$. But in this case you might be able to find a more specific direct argument, if you look for it

But this argument in fact works not just for $$E$$ but for any completely regular $$T_1$$ space: these all embed into a power of reals.

• In the part for showing T has a continuous inverse, is the equivalence of it to that other fact a standard result? Otherwise, I am having difficulty giving a direct proof for T having a continuous inverse~ – Homaniac Mar 6 at 18:00
• @Homaniac it’s a standard result in some books (Engelking general topology ) and in my old university notes. I’ve proved it on this site too. – Henno Brandsma Mar 6 at 18:08
• Thanks I see, do you have any links? I'll try to look it up~ – Homaniac Mar 6 at 18:13
• @Homaniac 34.2 in Munkres is one place. – Henno Brandsma Mar 6 at 18:40
• @Homaniac this thread on another forum has the proof too. – Henno Brandsma Mar 6 at 18:45