Definition of canonical isomorphism between two vector spaces I'm reading "Linear Algebra via Exterior Products" by Winitzki. The part leading up to the canonical isomorphism definition firstly proves the result that an $n$-dimensional vector space $V$ over $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$. Suppose the isomorphism is defined as $\hat{M}v = (x_1, \ldots, x_n)$, where the coordinates $x_i$ are given in the standard basis $\{e_i\}$.
Then it says:

Note that the isomorphism $\hat{M}$ will depend on the choice of the
  basis: a different basis $\{e′_i\}$ yields a different map
  $\hat{M_1}$. For this reason, the isomorphism $\hat{M}$ is not
  canonical.
...
A linear map between two vector spaces $V$ and $W$ is canonically
  defined or canonical if it is defined independently of a choice of
  bases in $V$ and $W$.

I can't understand why changing the basis of $\mathbb{R}^n$ would yield a different linear map. Shouldn't the map remain the same? Only the representation of the output vectors would change because of the basis change. 
And by that logic, for any linear map, changing the basis would change the coordinates, and hence the representation, of the output vectors. So what exactly is meant by a linear map "defined independently of the choice of bases"?

Consider the vector space of linear and constant polynomials with real coefficients:
$$V=\{a+bx: a,b \in \mathbb R\}.$$
Let's also consider the bases $B_1=\{1,x\}$ and $B_2=\{x,x+1\}$. If you express in each basis the polynomial $f(x)=2-3x$ you get
$$f(x)=2\cdot 1+(-3)\cdot x$$
and
$$f(x)=(-5)\cdot x+2\cdot (x+1).$$
Then, if we consider the first basis the isomorphism $\hat M$ is such that
$$\hat M(2-3x)=(2,-3)$$
but if we take the second basis we get
$$\hat M(2-3x)=(-5,2).$$
So that when you say "let $B$ be a basis...", the choice of that basis determines which homeomorphism you're defining.
