For a compact Lie group $G$, consider the map $f : G^{2n} \to G$ given by $f(A_1, B_1, \ldots, A_n, B_n) = \displaystyle\prod_{i = 1}^{n} A_i B_i A_i^{-1} B_i^{-1}$
A theorem of Goldman (from the 'Fundamental Groups of Surfaces' paper) says that the rank of $f$ at a point $(A_1, B_1, \ldots, A_n, B_n)$ equals the codimension of the centralizer of $\{A_1, B_1, \ldots, A_n, B_n\}$.
Recall that a point $(A_1, B_1, \ldots, A_n, B_n) \in G^{2n}$ is called reducible if all $A_i$ and $B_j$ comute with each other, and irreducible otherwise. Then for the case $G = SU(2)$ (and apparently more generally for $U(n)$) the above theorem should imply that $f$ is a submersion exactly at irreducible points. But why is this?
Let $H \subset G$ be a closed subgroup. First: $C_G(H)$ is positive-dimensional if and only if the adjoint representation has nontrivial fixed points when restricted to $H$. To see this, consider the tangent space $\mathfrak{c_h} \subset \mathfrak g$; I claim that the action of $H$ preserves this. This is nothing particularly clever; if $h \in H$, then by definition $x \mapsto hxh^{-1}$ acts as the identity on $C_G(H)$, whence the adjoint representation acts as the identity on $\mathfrak{c_h}$. Conversely, suppose $H$ acts as the identity on some subspace $\mathfrak k$. By the Lie group-algebra correspondence, $\mathfrak k$ is the tangent space of a unique Lie subgroup $K \subset G$; because $hxh^{-1}$ fixes $\mathfrak k$, it must at least preserve the subgroup $K$; invoking the correspondence again, since there is a unique map corresponding to the induced map $\mathfrak h \to \mathfrak k$, it must be the constant map. So $K \subset C_G(H)$ as desired.
(This is all quite natural from the perspective of differential geometry; if $\rho: \pi_1(M) \to G$ is reducible, we literally mean that $C_G(H)$ is larger than $Z(G)$, so that the set of gauge transformations fixing the flat connection $\rho$ is no larger than $Z(G)$; $C_G(\rho(G))$ is always the group of gauge transformations fixing $A_\rho$. When $C_G(H)$ is positive-dimensional, we have even better: there are nontrivial parallel sections of the flat bundle $\mathfrak g_\rho$, whereas this is false of irreducible representations.)
Now suppose $\{A_i, B_i\}$ act by the adjoint representation with a nontrivial fixed point space. Lifting this fixed point space (which must be 1-dimensional, as the only compact 2D Lie group is $T^2$, which is not a subgroup of $SU(2)$) we see that the $A_i, B_i$ fix some circle $S^1$ by conjugation. But elements of $SU(2)$ fix a circle under conjugation if and only if they are in that circle. (Exercise; let the circle be $S(U(1) \times U(1)) \subset SU(2)$ for convenience.) So the $A_i, B_i$ lie inside the abelian group $S^1$, so they all commute with one another. Conversely, if $A_i, B_i$ all commute with each other, they're simultaneously diagonalizable, so they lie in some conjugate of the diagonal subgroup of $SU(2)$ - which is an $S^1$.
It's harder to see, but still true, that the only closed subgroups that can occur as $C_G(H) \subset SU(2)$ are, up to conjugacy, the $\Bbb Z/2$, the circle subgroup, and $SU(2)$ itself.