Let $(V, ||\cdot||)$ be vector space over $\mathbb{R}$ normed and finite-dimensional $n$, $\Omega \subset V$ open and bounded, $\varphi: \overline{ \Omega} \to V$ continuous and $b \not\in \varphi(\partial \Omega)$. Fixing the bases $$\mathcal{B} = \{v_1, \ldots, v_n\} \ \ \mbox{e} \ \ \mathcal{C} = \{e_1, \ldots, e_n\}$$ of $V$ and $\mathbb{R}^n$, respectively, we can consider linear isomorphism $T: V \to \mathbb{R}^n$ satisfying $$T(v_i) = e_i,$$ for all $i \in \{1, \ldots, n\}$. Under these conditions, we can take the commutative diagram
$\require{AMScd}$ \begin{CD} \overline{\Omega} @>{\varphi}>> V\\ @V{T}VV @VV{T}V\\ T(\overline{\Omega}) @>>{f}> \mathbb{R}^n \end{CD}
that motivates us to define $$d_V(\varphi, \Omega, b) := d_{\mathbb{R}^n}(f, T(\Omega), T(b))$$ to be the topological degree of $\varphi$ over the vector space $V$ in $\Omega$ at point $b$.
Question: How to show that such a definition is independent of the base of $V$? In other words, for any other base $\mathcal{B'} = \{w_1, \ldots, w_n\}$ of $V$, in which the application is defined $S: V \to \mathbb{R} ^n$ that satisfies $$S(w_i) = e_i,$$ for all $i \in \{1, \ldots, n\}$, how to show that $$d_{\mathbb{R}^n}(f, T(\Omega), T(b)) = d_{\mathbb{R}^n}(S \circ \varphi \circ S^{-1 }|_{S(\overline{\Omega})}, S(\Omega), S(b))?$$