= Solution
A real <Lie group> is a group $G$ equipped with a finite-dimensional real <smooth manifold> structure, conventionally Hausdorff and second countable, for which multiplication $G\times G\to G$ and inversion $G\to G$ are smooth.
Its <tangent space> at the identity $e$ is defined by smooth curves through $e$: two curves represent the same tangent vector when their derivatives in a local chart agree at zero. For $X\in T_eG$, left translation $L_g(h)=gh$ determines the <left-invariant vector field>
$$
X^L(g)=(dL_g)_eX.
$$
The commutator of these derivations on smooth functions is another <left-invariant vector field>, so define the <Lie bracket>
$$
\boxed{[X,Y]=[X^L,Y^L](e),\qquad\operatorname{Lie}(G)=T_eG.}
$$
This construction supplies the <Lie algebra> structure. In a <Matrix Lie group>, $X^L(g)=gX$ and differentiating the two fields gives $[X,Y]=XY-YX$.
For the <special linear group>, differentiate the <determinant> along a curve $g(t)$ through $I$. Expansion of the <determinant>, or its differential $D\det_A(H)=\det(A)\operatorname{tr}(A^{-1}H)$, gives
$$
\left.\frac d{dt}\det(I+tX)\right|_{t=0}=\operatorname{tr}X.
$$
The <determinant> differential is surjective at $I$, so the level set $\det=1$ has <tangent space> equal to its differential's <kernel>. Equivalently, every tangent matrix has zero <trace>, and every zero-trace matrix supplies a curve $e^{tX}$ of <determinant> $e^{t\operatorname{tr}X}=1$. Hence
$$
\boxed{\operatorname{Lie}(\mathrm{SL}_n(\mathbb R))=\mathfrak{sl}_n(\mathbb R)=\{X:\operatorname{tr}X=0\},}
$$
with the matrix <commutator> bracket. The same calculation over $\mathbb C$ gives the complex <special linear Lie algebra> $\mathfrak{sl}_n(\mathbb C)$; as a real group, the complex group has that space viewed as a real <Lie algebra>.
The <matrix exponential> is the everywhere-convergent series
$$
\boxed{\exp X=\sum_{k=0}^\infty\frac{X^k}{k!},\qquad X\in\mathfrak{gl}_n,}
$$
whose inverse matrix is $\exp(-X)$. The <matrix logarithm> is locally defined near $I$ by
$$
\boxed{\log(I+A)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}kA^k,\qquad\|A\|<1.}
$$
These maps are inverse on suitable neighbourhoods of zero and $I$, giving a <logarithmic chart of a matrix Lie group>. A logarithm is not a globally single-valued inverse of the exponential.
Every invertible complex matrix nevertheless has at least one <matrix logarithm>. Put it in <Jordan normal form>. For a block $J=\lambda I+N$, $\lambda\ne0$ and $N^s=0$, choose any complex scalar logarithm $\ell$ of $\lambda$ and set
$$
L=\ell I+\sum_{k=1}^{s-1}\frac{(-1)^{k+1}}k\left(\frac N\lambda\right)^k.
$$
The finite logarithm and exponential identities in the nilpotent variable give $e^L=\lambda(I+N/\lambda)=J$. Combine the blocks and conjugate back. Thus \b[the exponential map is surjective on] $\mathrm{GL}_n(\mathbb C)$, by <existence of a logarithm for every invertible complex matrix>.
A connected counterexample is $\mathrm{SL}_2(\mathbb R)$. It is connected: <polar decomposition of an invertible real matrix> writes each element as $KP$, with $K\in\mathrm{SO}(2)$ and $P$ positive definite symmetric of <determinant> one; $\mathrm{SO}(2)$ is connected and $P$ is connected to $I$ through $P^t$.
But $g=\operatorname{diag}(-2,-1/2)$ belongs to this group and is not $e^X$ for any real $X$. Such an $X$ would commute with $e^X=g$. Since $g$ has two distinct real <eigenvalues>, direct commutation makes $X$ a real <diagonal matrix>. Its exponential has positive diagonal entries, a contradiction. Therefore
$$
\boxed{\exp:\mathfrak{sl}_2(\mathbb R)\longrightarrow\mathrm{SL}_2(\mathbb R)\text{ is not surjective}.}
$$
This is an <exponential-surjectivity obstruction from distinct negative eigenvalues>; connectedness does not eliminate the obstruction.
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