= Solution
A <Lie group> is a <group> that is also a <smooth manifold>, with <differentiable> multiplication and inversion. A <Lie algebra> is a <vector space> with a bilinear alternating <Lie bracket> satisfying the <Jacobi identity>.
For a <Matrix Lie group> $G$, put $\mathfrak g=T_I G$. If $X,Y\in\mathfrak g$, the matrix <commutator> $[X,Y]=XY-YX$ again lies in $\mathfrak g$: the group commutator $e^{tX}e^{sY}e^{-tX}e^{-sY}$ lies in $G$, and the coefficient of $ts$ in its <matrix logarithm> is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_I G$, with the commutator bracket, is the <Lie algebra of a matrix Lie group> $\mathcal L(G)$.
The <special linear group> $SL(2,\mathbb R)$ is the inverse image of the regular value $1$ under the smooth <determinant> map, and multiplication and inversion are smooth. Differentiating $\det(I+tA)=1+t\operatorname{tr}A+O(t^2)$ shows that
$$
\mathcal L(SL(2,\mathbb R))=\mathfrak{sl}_2(\mathbb R)
=\left\{\begin{pmatrix}a&b\\c&-a\end{pmatrix}:a,b,c\in\mathbb R\right\}.
$$
The <Cayley-Hamilton theorem> applied to a trace-zero two-by-two matrix gives
$$
\boxed{A^2=-\det(A)I_2.}
$$
The <exponential map of a matrix Lie group> is the <matrix exponential>
$$
\operatorname{Exp}(A)=e^A=\sum_{n=0}^{\infty}\frac{A^n}{n!}.
$$
Since $\det(e^A)=e^{\operatorname{tr}A}=1$, its image lies in $SL(2,\mathbb R)$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt d$,
$$
e^A=\cos r\,I+\frac{\sin r}{r}A,
\qquad \operatorname{tr}(e^A)=2\cos r\geq-2.
$$
If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,
$$
e^A=\cosh r\,I+\frac{\sinh r}{r}A,
\qquad \operatorname{tr}(e^A)=2\cosh r\geq2.
$$
Hence
$$
\boxed{\operatorname{tr}(\operatorname{Exp}A)\geq-2.}
$$
But $\operatorname{diag}(-2,-1/2)\in SL(2,\mathbb R)$ has trace $-5/2$. It is therefore outside the image, so \b[the exponential map is not surjective].
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