Isomorphisms of the cantor set Since there exists a function from the cantor set $\mathcal{C}$ to $[0,1]$ which I will call $f$, and they're also exists a function from $[0,1]$ to $\mathbb{R}$ which I will call $g$, then the following could be said, $h=f\circ g$ and hence, $\exists h:\mathcal{C}\to\mathbb{R}$. Now assuming that $\exists h^{-1}:\mathbb{R}\to\mathcal{C}$. Could one say the following is true?
$$\mathcal{C}\cong\mathbb{R}$$
Or in other words, "there exists an isomorphism between the cantor set and the set of real numbers". Or even so far as to say there exists a homeomorphism between $\mathcal{C}$ and $\mathbb{R}$.
And if so, how do you prove it?
 A: Isomorphism : isos "equal", and "morphe "structure"
To talk about isomorphism we need two mathematical structure like sets (bijection ), well-ordeded sets( order isomorphism) groups, rings, fields, vector spaces ,metric spaces(isometry),topological spaces (homeomorphism) etc.


Since there exists a function from the cantor set $\mathcal{C}$ to $[0,1] $
which I will call $f$

You can choose such map $f$ bijective.

They're also exists a function from $[0,1] $ to $\Bbb{ R }$ which I
will call $g$

You can choose such map $g$ bijective.

Then the following could be said, $h=f\circ g $ and hence, $\exists
 h:\mathcal{C}\to \Bbb{R}$. Now assuming that $\exists
 h^{-1}:\Bbb{R}\to \mathcal{C}$.

Of course you can choose such map $h$ which is also bijective.

"there exists an isomorphism between the cantor set and the set of
real numbers"

Conclusion ( merely as sets) : The Cantor set $\mathcal{C}$ and $\Bbb{R}$ are isomorphic as sets( here isomorphism is determined by existence of a bijection between those two sets, known as equivalent or equipotent. In other words both sets have the same cardinality)



Since there exists a function from the cantor set $\mathcal{C}$ to $[0,1] $
which I will call $f$

We know every compact metric space is the homeomorphic image of the Cantor set i.e there is a continuous onto map  from the Cantor set to  $[0, 1]$ .
So you can choose $f$ continuous and onto but such $f$ can't be one-to-one. $($ Homeomorphism respect topological properties! Cantor set is not connected in fact totally disconnected whereas $[0, 1]$ is connected.$)$

They're also exists a function from $[0,1] $ to $\Bbb{ R }$ which I
will call $g$

Again $g$ can't be continuous bijection. $(\Bbb{R}$ is not compact whereas $[0, 1]$ is compact.$) $

Or even so far as to say there exists a homeomorphism between $\mathcal{C}$ and $\Bbb{R}$

It is clear that the function $h=f\circ g$ is not a homeomorphism. But there may be other function which can be a homeomorphism. But this is not possible! There is no homeomorphism between  $\mathcal{C}$ and $\Bbb{R}$.
Homeomorphism is a topological isomorphism ( bijection preserve set theoretic structure and bi-continuity preserve topological structure like compactness, connectedness, connected components, path components etc.)
The Cantor set and $\Bbb{R}$ has many conflicting topological properties.
