TITLE: Canonical Topology
QUESTION [6 upvotes]: Let $V$ be an n-dimensional vector space over $\mathbb{R}$.
Of course $V \cong \mathbb{R^n}$ since we can define an isomorphism
$$f:V\longrightarrow \mathbb{R^n}$$
by mapping basis elements to basis elements.
But such isomorphism requires a choice of basis.
I can define a topology $\tau = \{f^{-1}(V); V \ $is open in$\ \mathbb{R^n}\}$
But is $\tau$ a canonical topology? 
An isomorphism is canonical if it is defined without having to choose a basis. 
But what does canonical mean in terms topology? 
In my case, I had to "choose" a basis to define an isomorphism, then use that isomorphism to define open sets. So I'm not sure if it works...

REPLY [10 votes]: It is a canonical topology in the following two senses:

A priori, the topology $\tau$ depends on the choice of an isomorphism $f \colon V \rightarrow \mathbb{R}^n$ so let's denote $\tau$ by $\tau_f$. However, if you pick a different linear isomorphism $g \colon V \rightarrow \mathbb{R}^n$ then you have $\tau_f = \tau_g$ so in fact, the topology doesn't depend on the choice of an isomorphism. To see that, note that we can find a linear isomorphism $T \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $T \circ f = g$. Since $T$ is a linear isomorphism, it is, in particular, a homeomorphism of $\mathbb{R}^n$ (with the standard topology) and so a set $U$ is open in $\mathbb{R}^n$ iff $T^{-1}(U)$ is open in $\mathbb{R}^n$. But since $g^{-1}(U) = f^{-1}(T^{-1}(U))$, the resulting topologies are the same.
You can characterize $\tau$ without any choice by saying that $\tau$ is the weakest topology on $V$ with respect to which, any linear map $T \colon V \rightarrow \mathbb{R}^n$ (or even any linear map $T \colon V \rightarrow \mathbb{R}^m$) is continuous (where the right hand side gets the usual topology). Thus
$$ \tau = \{ T^{-1}(U) \, | U \subseteq \mathbb{R}^n \textrm{ is open}, T \colon V \rightarrow \mathbb{R}^n \textrm{ is a linear isomorphism}\} \\
= \{ T^{-1}(U) \, | \, U \subseteq \mathbb{R}^n \textrm{ is open}, T \colon V \rightarrow \mathbb{R}^n \textrm{ is linear} \} \\
= \{ T^{-1}(U) \, | \, U \subseteq \mathbb{R}^m \textrm{ is open}, T \colon V \rightarrow \mathbb{R}^m \textrm{ is linear} \} $$
and you eliminate any "choice" by taking "all the possible choices".
To see that the three definitions for $\tau$ coincide, let $T \colon V \rightarrow \mathbb{R}^m$ be a linear map and let $U \subseteq \mathbb{R}^m$ be an open set. We need to show that we can find an open $\tilde{U} \subseteq \mathbb{R}^n$ and a linear isomorphism $\phi \colon V \rightarrow \mathbb{R}^n$ such that $T^{-1}(U) = \phi^{-1}(\tilde{U})$. Choose some linear isomorphism $\phi \colon V \rightarrow \mathbb{R}^n$ and write $T = (T \circ \phi^{-1}) \circ \phi$ where $T \circ \phi^{-1} \colon \mathbb{R}^n \rightarrow \mathbb{R}^m$ is linear, hence continuous. Then
$$ T^{-1}(U) = \phi^{-1} \left( \underbrace{\left( T \circ \phi^{-1} \right)^{-1} (U)}_{\tilde{U}} \right). $$