The Lie algebra version of $1 \to SU(N) \to U(N) \to U(1) \to 1$ For the unitary group $U(N)$ and the special unitary group $SU(N)$, we have the exact sequence
$$
1 \to SU(N) \to U(N) \to U(1) \to 1.
$$
How does this behave at the level of algebras? I suppose we still get an exact sequence of Lie algebras?
In particular, I would be very interested in understanding this sequence in terms of the Gell-Mann matrix 
https://en.wikipedia.org/wiki/Gell-Mann_matrices
presentation of $\frak{su}_3$:
https://en.wikipedia.org/wiki/Special_unitary_group#Lie_algebra
 A: By using the exponential map and  the one-parameter subgroups of these Lie groups, one does indeed get an exact sequence
$$0\to\mathfrak{su}(N)\to\mathfrak{u}(N)\to\mathfrak{u}(1)\to 0.$$
See Exactness of Lie algebra exact sequence.  (This is true in some more generality: in the comments of Exactness of the lie algebra functor there is a reference to Theorem 4.20 of the book The Lie Theory of Connected Pro-Lie Groups which proves this for what they call a pro-Lie group, a topological group that is isomorphic algebraically and topologically to a subgroup of a product of Lie groups.)
At the level of algebras, recall that when there is a Lie group homomorphism $f:G\to H$, then there is a commuting square
$$\require{AMScd}
\begin{CD}
\mathfrak{g} @>{df_e}>> \mathfrak{h}\\
@V{\exp}VV  @VV{\exp}V\\
G @>>f> H
\end{CD}$$
So, we need only take the derivatives of the maps in the exact sequence at the identity.  The derivative of the inclusion map $SU(N)\to U(N)$ is simply the inclusion map $\mathfrak{su}(N)\to\mathfrak{u}(N)$.  The derivative of $\det:U(n)\to U(1)$ comes from Jacobi's formula.  Hence, we have the commutative diagram
$$\require{AMScd}
\begin{CD}
\mathfrak{su}(N) @>>> \mathfrak{u}(N) @>{\operatorname{tr}}>> \mathfrak{u}(1)\\
@V{\exp}VV @V{\exp}VV  @VV{\exp}V\\
SU(N) @>>> U(N) @>>{\operatorname{det}}> U(1)
\end{CD}$$
Notice that the top row is stating that $\mathfrak{su}(N)$ consists of the traceless matrices from $\mathfrak{u}(N)$.
Using the convention that $\mathfrak{u}(N)$ is $N\times N$ Hermitian matrices using $[A,B]:=-i(AB-BA)$ as its Lie bracket (which I believe is the convention being used for the Gell-Mann matrices), then introducing the identity matrix to the eight Gell-Mann matrices gives the last generator for $\mathfrak{u}(3)$, and the structure coefficients for the bracket with the identity matrix are all $0$.
The choice of the identity matrix was arbitrary -- it could have been any matrix with nonzero trace.  It so happens that this choice induces a splitting $\mathfrak{u}(N)\approx\mathfrak{su}(N)\oplus\mathfrak{u}(1)$.  Given $A\in\mathfrak{u}(N)$, let $A_0=A-\frac{1}{N}\operatorname{tr}(A)I$ and let $A_1=\frac{1}{N}\operatorname{tr}(A)I$.  Then obviously $A=A_0+A_1$ and $\operatorname{tr}(A_0)=0$.  Furthermore, if $A$ is in $\mathfrak{su}(N)$, then $A=A_0$ and $A_1=0$.  One can check that $A\mapsto A_0$ preserves the Lie bracket.
