= Solution
The <Lie algebra> of a <Lie group> $G$ is the vector space $\mathfrak g=T_eG$ equipped with the bracket transferred from <left-invariant vector fields>. For $a\in G$, write $L_a(g)=ag$. A field is left-invariant when $(D_gL_a)X_g=X_{ag}$ for all $a,g$. Given $v\in T_eG$, define
$$
X_v(g)=(D_eL_g)v.
$$
This is smooth and left-invariant, because $L_aL_g=L_{ag}$ and the <chain rule> applies. Conversely a left-invariant field is determined by its value at $e$, through this formula. Thus
$$
\boxed{v\longmapsto X_v}
$$
is a vector-space isomorphism, whose inverse is evaluation at the identity.
The <Lie bracket of vector fields> is preserved by <diffeomorphisms>: on functions this follows by transporting the commutator of derivations. Therefore the bracket of two left-invariant fields is again left-invariant. Define $[v,w]=[X_v,X_w]_e$. Bilinearity, antisymmetry and the <Jacobi identity> follow from the same properties of commutators of derivations. This supplies the asserted <Lie algebra> structure.
For the <general linear group> $GL_n(\mathbb R)$, invertible matrices form an open subset of $M_n(\mathbb R)$, so $T_IGL_n(\mathbb R)=M_n(\mathbb R)$. The left-invariant field determined by $A$ is $X_A(g)=gA$. Its ambient derivative is $D_gX_A(H)=HA$. Therefore <differentiating left-invariant matrix fields> gives
$$
[X_A,X_B](g)=D_gX_B(gA)-D_gX_A(gB)=g(AB-BA).
$$
Hence
$$
\boxed{\mathfrak{gl}_n(\mathbb R)=M_n(\mathbb R),\qquad[A,B]=AB-BA.}
$$
This is the <general linear Lie algebra>, with the ordinary <matrix commutator>; right invariance with the same identification would give the opposite sign.
Now let $F:G\to H$ be a smooth <Lie group homomorphism>. It sends the identity to the identity and induces the linear map
$$
F_*=D_eF:T_eG\to T_eH.
$$
Since $F\circ L_g=L_{F(g)}\circ F$, differentiation gives
$$
(D_gF)X_v(g)=X_{F_*v}(F(g)).
$$
These fields are $F$-related; no injectivity or surjectivity of $F$ is needed. For every smooth function $f$ on $H$,
$$
X_v(f\circ F)=(X_{F_*v}f)\circ F.
$$
Apply this first with $w$ and then with $v$, and subtract the reversed order. The result is
$$
[X_v,X_w](f\circ F)=([X_{F_*v},X_{F_*w}]f)\circ F.
$$
At $e$, smooth functions detect tangent vectors, so
$$
\boxed{F_*[v,w]=[F_*v,F_*w].}
$$
This proves that the <differential of a Lie group homomorphism preserves Lie brackets>, and hence that $F_*$ is a <Lie algebra homomorphism>.
For the exponential identity, let $\eta_v$ be the integral curve of $X_v$ through $e$. Left invariance and uniqueness show $\eta_v(t+s)=\eta_v(t)\eta_v(s)$ for small times: translating the curve through $e$ gives the curve through $\eta_v(s)$. Repeating this identity extends the curve to all real times. This proves <completeness of left-invariant vector fields> without assuming that an arbitrary smooth field is complete. Rescaling the parameter also gives $\eta_v(t)=\eta_{tv}(1)$. By the definition of the <Exponential map of a Lie group>, $\exp_G(v)=\eta_v(1)$.
The field-related identity shows that $F(\eta_v(t))$ is an integral curve of $X_{F_*v}$ starting at $e_H$. By uniqueness it equals $\eta_{F_*v}(t)$ for all times. At time one,
$$
\boxed{F(\exp_Gv)=\exp_H(F_*v).}
$$
This is the <naturality of the Lie group exponential>.
Apply this to the <determinant> homomorphism $GL_n(\mathbb R)\to\mathbb R^\times$. The derivative of the determinant at $I$ is
$$
D_I\det(A)=\operatorname{tr}A,
$$
since the permutation formula gives $\det(I+tA)=1+t\operatorname{tr}A+O(t^2)$. The left-invariant field on $\mathbb R^\times$ with initial tangent $c$ is $y\mapsto cy$; its curve through one solves $y'=cy$, hence is $e^{ct}$. The <matrix exponential determinant identity> therefore follows from naturality:
$$
\boxed{\det(\exp A)=e^{\operatorname{tr}A}.}
$$
This argument uses only the flow definition of the exponential and the scalar exponential, not an unproved matrix formula.
To identify it explicitly with the usual <matrix exponential>, the series $E(t)=\sum_{k\geq0}t^kA^k/k!$ converges in <operator norm>, uniformly with its differentiated series on bounded intervals. Termwise differentiation gives $E'=EA$ and $E(0)=I$. The series commutes with $A$, and differentiating $E(t)E(-t)$ gives zero, so $E(t)E(-t)=I$. Thus $E(t)$ stays in $GL_n(\mathbb R)$ and, by uniqueness, is the integral curve of $X_A$ through $I$. \b[The matrix series is exactly the flow-defined exponential], justifying either notation in the determinant identity.
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