The exponential map is surjective on every connected Lie group

The open matrix group $G$ is connected. Indeed, [F4] writes every $B\in G$ as $B=SU$ with $U$ positive definite and $S\in \mathrm{SO}(2)$. The path $(1-t)U+tI$ stays positive definite, and every $S\in \mathrm{SO}(2)$ is a rotation joined to $I$ by varying its angle. Thus $B$ is path connected to $I$. Also $\det A=1$, so $A\in G$; explicitly $t\mapsto R(\pi t)\mathrm{diag}(2^t,2^{-t})$ joins $I$ to $A$.

For a real matrix $X$, the absolutely convergent series $E(t)={\sum }_{n\ge 0}{t}^n{X}^n\mathrm{/}n!$ solves $E^{\mathrm{\prime }}=XE$ and $E(0)=I$. By [F1] and [F3], uniqueness identifies $E(1)$ with ${\exp }_{G}X$. In particular, $X$ commutes with ${\exp }_{G}X$.

If ${\exp }_{G}X=A$, step 1.2 says $XA=AX$. Because $A$ has two distinct real eigenvalues, this equation forces $X$ to preserve each coordinate line and hence to be diagonal, say $X=\mathrm{diag}(u,v)$. Then ${\exp }_{G}X=\mathrm{diag}(e^u,e^v)$ has positive diagonal entries, contradicting the two negative entries of $A$.

Thus $A$ is not exponential although $G$ is nonempty, connected, and four-dimensional. The determinant is nonzero and no degeneracy is hidden. The paths include both endpoints. The Euclidean inner product in [F4] is an explicit finite-dimensional witness and invokes no metric-existence theorem. Countable choice is assumed exactly to use [F1]; no additional choice or biconditional occurs, and zero and one dimensions cannot invalidate this explicit counterexample.