# Why can a ZY decomposition not decompose an arbitrary single qubit gate?

To decompose an arbitrary single qubit gate, we need to do a "$$e^{i \alpha}$$ZYZ" decomposition.

Intuitively however I do not understand why a "$$e^{i \alpha}$$ZY" decomposition is not sufficient:

An arbitrary single qubit state can be decomposed into its $$|0\rangle$$ and $$|1\rangle$$ components. We can move both $$|0\rangle$$ and $$|1\rangle$$ to any arbitrary state $$|\psi\rangle$$ on the Bloch sphere by first applying a $$Y$$ rotation by some angle $$\gamma$$ and then a $$Z$$ rotation by some angle $$\beta$$.

You've shown that you can put the state $$|0\rangle$$ wherever you want. But you also need to put the $$|+\rangle$$ state in the correct location on the great circle of states perpendicular to where $$|0\rangle$$ ended up. This is the third degree of freedom in the problem, requiring a third degree of freedom in the solution.
As you rightly say, a sequence $$e^{i\alpha}ZY$$ can prepare any single-qubit pure state. That is not the same as having an arbitrary single qubit unitary.
A unitary can be defined by its action on the basis states. So, you've already got arbitrary action on the $$|0\rangle$$ input, which is a great start. Moreover, the unitary must take $$|1\rangle$$ to a state orthogonal to the one that $$|0\rangle$$ is mapped to, which is incredibly limiting. However, there is still a global phase factor allowed in the possible definition of $$U|1\rangle$$ (it's an important global factor because when applied to an input in superposition, it's a relative phase between the two components). How do you incorporate that extra phase? Put an arbitrary phase rotation in first to add that phase to the $$|1\rangle$$ component.