The proof constructs an explicit surjective homomorphism $\Phi$ from $\mathrm{GL}_2(\mathbb{C})$ to $\mathcal{M}$, computes its kernel, and applies the [First Isomorphism Theorem](/theorems/791). Then the $\mathrm{SL}_2$ version follows by restricting and normalising. **Step 1: Construct the homomorphism.** Define: \begin{align*} \Phi : \mathrm{GL}_2(\mathbb{C}) &\to \mathcal{M} \\ \begin{pmatrix} a & b \\ c & d \end{pmatrix} &\mapsto f(z) = \frac{az + b}{cz + d}. \end{align*} **Step 2: $\Phi$ is a homomorphism.** The composition formula from the proof of the [Möbius Group](/theorems/808) theorem shows that $\Phi(AB) = \Phi(A) \circ \Phi(B)$ for all $A, B \in \mathrm{GL}_2(\mathbb{C})$: the coefficients of the composed Möbius map are exactly the entries of the matrix product. **Step 3: $\Phi$ is surjective.** Every Möbius map $\frac{az+b}{cz+d}$ with $ad - bc \neq 0$ is the image of $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{GL}_2(\mathbb{C})$. **Step 4: Compute the kernel.** $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \ker(\Phi)$ iff $\frac{az + b}{cz + d} = z$ for all $z \in \mathbb{C}_\infty$. Testing $z = \infty$ gives $c = 0$; $z = 0$ gives $b = 0$; $z = 1$ gives $a = d$. So $\ker(\Phi) = \left\{\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} : \lambda \in \mathbb{C} \setminus \{0\}\right\} = Z(\mathrm{GL}_2(\mathbb{C}))$. By the [First Isomorphism Theorem](/theorems/791): $\mathrm{GL}_2(\mathbb{C}) / Z(\mathrm{GL}_2(\mathbb{C})) \cong \mathcal{M}$. **Step 5: The $\mathrm{SL}_2$ version.** Restrict $\Phi$ to $\mathrm{SL}_2(\mathbb{C})$. This is still surjective: given any $f(z) = \frac{az+b}{cz+d}$, the matrix $\frac{1}{\sqrt{ad-bc}}\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ has determinant $1$ and maps to the same Möbius transformation. The kernel is $\ker(\Phi) \cap \mathrm{SL}_2(\mathbb{C}) = \{\lambda I : \lambda^2 = 1\} = \{\pm I\}$. By the [First Isomorphism Theorem](/theorems/791): $\mathrm{SL}_2(\mathbb{C}) / \{\pm I\} \cong \mathcal{M}$.