Inspired by this question About noncommutative sphere and inspired by the fact that the classical sphere is the one point compactificatiin of $\mathbb{R}^2$ we ask the question below:
Is the non commutative sphere $A_q$ mentioned in the above linked post a (minimal)unitization of a non commutative deformation $B_q$ of $C_0(\mathbb{R}^2)$?
One may pose this question for any other NC deformation of classical sphere
This is really a long comment. We first show that the universal abelian $C^\ast$-algebra $A$ generated by two generators $a,b$ subject to the relations $a = a^\ast$ and $|b|^2 = a(1-a)$ is $C_0(\mathbb{R}^2)$, with $a = \frac{1}{1+x^2+y^2}$ and $b = \frac{x+iy}{1+x^2+y^2}$.
To do this, we consider the unitization $\tilde{A}$ of $A$. By considering the universal property satisfied by $\tilde{A}$, it is clear that it is the universal unital abelian $C^\ast$-algebra $A$ generated by two generators $a,b$ subject to the relations $a = a^\ast$ and $|b|^2 = a(1-a)$. Note that $a(1-a) = |b|^2 \geq 0$, so as $a$ is self-adjoint, by functional calculus, we have $0 \leq a \leq 1$. Then,
$$\|b\|^2 = \||b|^2\| = \|a(1-a)\| \leq \sup_{0 \leq x \leq 1} |x(1-x)| = \frac{1}{4}$$
So $\|b\| \leq \frac{1}{2}$. Let $c = \text{Re}(b)$ and $d = \text{Im}(b)$. Then $c,d$ are both self-adjoint elements of norm at most $\frac{1}{2}$. We renormalize these a bit by defining $u = 2a-1$, $v = 2c$, $w = 2d$. Then $\tilde{A}$ is generated as a unital $C^\ast$-algebra by $u,v,w$. We also have $\sigma(u), \sigma(v), \sigma(w) \subset [-1, 1]$, so by functional calculus, there is a surjective $\ast$-homomorphism:
$$\pi: B := C([-1, 1]^3) \to \tilde{A}, \pi(f \otimes g \otimes h) = f(u)g(v)h(w)$$
Let the three coordinate projections on $[-1, 1]^3$ be denoted by $x,y,z$, respectively. Then $\pi(x) = u$, $\pi(y) = v$, $\pi(z) = w$. As $\tilde{A}$ is a quotient of $B$, $\tilde{A}$ is the algebra of continuous functions on some closed subset $X \subset [-1, 1]^3$. The relation defining $\tilde{A}$ is $|b|^2 = a(1-a)$, which, translated in terms of $u,v,w$, is,
$$(\frac{v}{2})^2 + (\frac{w}{2})^2 = \frac{1+u}{2} \cdot \frac{1-u}{2} \Leftrightarrow u^2 + v^2 + w^2 = 1$$
So $\tilde{A} = B/\langle x^2 + y^2 + z^2 - 1 \rangle$. Thus, $\tilde{A} = C(X)$ where,
$$X = \{(x,y,z) \in [-1, 1]^3: x^2 + y^2 + z^2 = 1\}$$
That is, $\tilde{A} = C(S^2)$. Note that $u,v,w$ corresponds to the three Cartesian coordinate functions $x,y,z$, so,
$$a = \frac{1+u}{2} = \frac{1+x}{2}, b = \frac{v+iw}{2} = \frac{y+iz}{2}$$
Now, $A$ is the (non-unital) $C^\ast$-subalgebra generated by $a$ and $b$. The only point on the unit sphere where both $a$ and $b$ are zero is $(-1,0,0)$, so,
$$A = C_0(S^2 \setminus \{(-1,0,0)\}) \simeq C_0(\mathbb{R}^2)$$
Finally, we show that we may choose $a = \frac{1}{1+x^2+y^2}$ and $b = \frac{x+iy}{1+x^2+y^2}$ in $C_0(\mathbb{R}^2)$. This follows from choosing an appropriate homeomorphism $\phi: \mathbb{R}^2 \to S^2 \setminus \{(-1,0,0)\}$:
$$\phi(x,y) = (\frac{1-x^2-y^2}{1+x^2+y^2}, \frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2+y^2})$$
It is a homeomorphism as, one may easily verify, it admits a continuous inverse,
$$\phi^{-1}(x,y,z) = (\frac{y}{1+x}, \frac{z}{1+x})$$
Thus, under this identification, $a$, as an element of $C_0(\mathbb{R}^2)$, is,
$$\frac{1+x}{2} \circ \phi = \frac{1}{1+x^2+y^2}$$
And $b$ is,
$$\frac{y+iz}{2} \circ \phi = \frac{x+iy}{1+x^2+y^2}$$
This proves the claim.
Now, as I have observed in comments, the relations defining $A$ may be easily restated as $A$ being the universal $C^\ast$-algebra generated by two generators $a,b$ subject to the relations:
$$a = a^\ast, ab = ba, ab^\ast = b^\ast a, bb^\ast = b^\ast b = a(1-a)$$
If we introduce a parameter $0 < q <1$ and deform these relations, we can then get the relations defining the quantum sphere modulo unitality,
$$a = a^\ast, ab = q^2ba, ab^\ast = q^{-2}b^\ast a, bb^\ast = q^{-2}a(1-a), b^\ast b = a(1-q^2a)$$
We retrieve the relations defining $A$ as $q \to 1$. Let the universal (non-unital) $C^\ast$-algebra generated by $a,b$ subject to the above deformed relations be $B_q$. Then $B_q$ is a deformation of $A = C_0(\mathbb{R}^2)$ and the unitization of $B_q$ is the quantum sphere, $A_q$.
