Canonical Topology 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...
 A: 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
$$ T^{-1}(U) = \phi^{-1} \left( \underbrace{\left( T \circ \phi^{-1} \right)^{-1} (U)}_{\tilde{U}} \right). $$
A: You may check that your topology does not depend on the choice of $f$: if you pick another basis, the open subsets will be the same.
This topology is canonical in the following sense. First, note that it makes $V$ into a topological vector space: the addition of vectors $V\times V\to V$ and the action $\mathbb{R}\times V\to V$ are continuous with respect to this topology. Then, any finite dimensional Hausdorff topological vector space necessarily has the topology that you described.
See an exposition in the blog of Terence Tao.
