Let $\mathcal S = \{A \in M_n(\mathbb R) : \rho(A) < 1\}$ where $\rho(\cdot)$ denotes the spectral radius. Let $f: \mathcal S \to M_n(\mathbb R)$ be given by $$ A \mapsto (A-I)^{-1}(A+I).$$ It can be shown the eigenvalues of the image $f(A)$ have negative real part. Now let us denote $\mathcal T = \{A \in M_n(\mathbb R): \text{Re}(\lambda_i(A)) < 0, \text{ for } i=1, \dots, n\}$. Now consider $f$ as a map into $\mathcal T$, i.e., $f: \mathcal S \to \mathcal T$.
I am wondering whether this map is bijective. It seems bijective into its image $f(\mathcal S)$ since if $B = (A-I)^{-1}(A+I) \implies A= (B-I)^{-1}(B+I)$. But is it surgective onto $\mathcal T$?
All you need to do is to determine that if $B\in\cal T$ then $A=(B-I)^{-1}(B+I) \in\cal S$. This comes down to eigenvalues: if $\lambda$ is in the left half-plane, then $|(\lambda+1)/(\lambda-1)|<1$. This is true.