= Solution
For an irreducible $V$, <Schur lemma> makes the central element $-I$ act by a scalar $\varepsilon$. Since $(-I)^2=I$, $\varepsilon=\mathord\pm1$, and on $V\otimes V$ it acts by $\varepsilon^2=1$. Thus $\{\mathord\pm I\}$ lies in the kernel and the action factors through $SU_2/\{\mathord\pm I\}$, as in <central parity on SU2 tensor products>.
This fails for reducible representations. Take $V=V_0\oplus V_1$. On the summand $V_0\otimes V_1\subset V\otimes V$, the element $-I$ acts as $-1$, so the tensor-square action does not factor through the quotient.
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