Showing that any linear bijection $f:\mathbb{R}^{n}\to E$ is an isomorphism I am a bit confused about a corollary on Page 17 from Cartan's Differential Calculus regarding the statement that every linear bijection from $\mathbb{R}^{n}$ to a normed vector space $E$ is an isomorphism.
I haven't been able to find any information that helped solve this question from other posts on here.
The definition of an isomorphism in this context is given as follows (from Page 15):

DEFINITION: A mapping $f:E\to F$ (where $E$ and $F$ are two normed v.s.) is an isomorphism if:
(1) $f$ is linear and continuous;
(2) there exists a linear continuous mapping $F\to E$ such that $g\circ f=\text{id}_{E}$ (the identity mapping of $E$) and $f\circ g=\text{id}_{F}$.

The theorem that proceeds the corollary is that all norms on $\mathbb{R}^{n}$ are equivalent. The corollary is:

Corollary. If $E$ is a normed vector space then every linear bijective mapping $f:\mathbb{R}^{n}\to E$ is an isomorphism. (Indeed, if $\rho$ is a norm on $E$, then $\rho\circ f$ is a norm on $\mathbb{R}^{n}$; the latter defines the same topology as the Euclidean norm, and hence the result.)

I understand why $\rho\circ f$ is a norm on $\mathbb{R}^{n}$ and why this norm defines the same topology as the Euclidean norm from the previous theorem, but I am not sure how this is used to conclude that $f$ is an isomorphism.
There is a theorem in the book on Page 15 that states if $E$ and $F$ are Banach spaces and $f:E\to F$ is a linear continuous bijective operator, then $f^{-1}:F\to E$ is also continuous and $f$ is an isomorphism. But as far as I am aware, in the corollary above $E$ will not be a Banach space in general, so showing that $f$ is continuous will not be sufficient.
Can someone please explain how it can be concluded from the explanation in the corollary (or from another method if this is not possible) that $f$ is an isomorphism?
 A: Let $A$ be an open subset of $\Bbb R^n$. If $p\in A$, then there is some $r>0$ such that $\|x-p\|<r\implies x\in A$. Since the usual norm $\|\cdot\|$ and $\rho\circ f$ are equivalent norms in $\Bbb R^n$, there is some $s>0$ such that$$\rho\bigl(f(x)-f(p)\bigr)<s\implies\|x-p\|<r\implies x\in A.$$But then $f(A)$ is an open subset of $E$: for each of its points, there is an open ball centered at that point which is contained in it. So, $f^{-1}$ is continuous: the inverse image of an open set is open.
A: There are some basic facts which you can easily verify:

*

*If $f : E \to F$ is a linear bijection between vector spaces, then the set theoretic inverse $f^{-1} : F  \to E$ is also a linear bijection. That is, $f$ is a vector space isomorphism.


*If $f : E \to F$ and $g : F \to G$ are isomorphisms of normed spaces, then $g \circ f$ is an isomorphism.


*If two norms $n, n'$ on a vector space $E$ are equivalent, then $id_E : (E,n) \to (E,n')$ is an isomorphism.
You know that $\rho' = \rho\circ f$ is a norm on $\mathbb{R}^{n}$ and thus $\rho'$ is equivalent to the Euclidean norm $n_e$ (i.e. this norm defines the same topology as the Euclidean norm). But now
$$f : (\mathbb R^n,\rho') \to (E,\rho)$$
by definition has the property that $\rho(f(x)) = \rho'(x)$ which shows that $f$ is continuous; note that $\rho (f(x) - f(y)) = \rho(f(x-y)) = \rho'(x-y)$.
Similar $f^{-1} : (E,\rho) \to (\mathbb R^n,\rho')$ is continuous. Hence $f : (\mathbb R^n,\rho') \to (E,\rho)$ is an isomorphism. But $id : (\mathbb R^n,n_e) \to (\mathbb R^n,\rho')$ is an isomorphism, hence $f = f \circ id :  (\mathbb R^n,n_e) \to (E,\rho)$ is one.
