Is $GL_2(\mathbb{C})\cong \hat{\mathbb{C}}\times \mathbb{C}\times \mathbb{C}^*\times \mathbb{C}^*$? Is $GL_2(\mathbb{C})\cong \hat{\mathbb{C}}\times \mathbb{C}\times \mathbb{C}^*\times \mathbb{C}^*$ ?
I found that $GL_2(\mathbb{C})/Z_2(\mathbb{C})\cong Mob$, where $Mob$ is the set of Möbius transformations, $Z_2(\mathbb{C})$ is the centre of $GL_2(\mathbb{C})$.
Therefore, $GL_2(\mathbb{C})\cong Z_2(\mathbb{C})\times Mob\cong Z_2(\mathbb{C})\times \left\{f:f=\dfrac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)},z_i\neq z_j, z_i\in\hat{\mathbb{C}}\right\}$
$\cong Z_2(\mathbb{C})\times \hat{\mathbb{C}}\times \mathbb{C}\times \mathbb{C}^*\cong\hat{\mathbb{C}}\times\mathbb{C}\times\mathbb{C}^*\times\mathbb{C}^*$
What if I changed $GL_2(\mathbb{C})$ into $GL_n(\mathbb{C})$, where $n\in\mathbb{N}^*$?
Could there be more general way to prove the case that $n=2$?

$\hat{\mathbb{C}}$ means $\mathbb{C}\cup\{\infty\}$.
 A: A group $G$ is an internal direct product $H\times K$ if every $g\in G$ is uniquely expressible as $g=hk$ and every $h\in H$ and $k\in K$ commute. Any direct product of abelian groups is abelian, so no nonabelian group can be a direct product of abelian groups. A weaker condition is if $G$ is an internal knit product $G=HK$ of two subgroups $H$ and $K$; this happens when every $g\in G$ is uniquely expressible as $g=hk$ for some $h\in H$ and $k\in K$, but with no assumption that $h$s and $k$s commute.
The QR factorization shows $\mathrm{GL}_n\mathbb{C}$ is a knit product of $\mathrm{U}(n)$ and the group $B$ of upper triangular matrices with real positive diagonal entries. This means $\mathrm{GL}_n\mathbb{C}\simeq \mathrm{U}(n)\times B$ is true topologically. Moreover, since positive reals are homeomorphic to all reals (via logarithms), $B\simeq\mathbb{R}^n\times\mathbb{C}^{\binom{n}{2}}\simeq\mathbb{R}^{n^2}$ topologically.
This is also related to the Iwasawa decomposition of a Lie group.
The unitary group $\mathrm{U}(n)$ is not so simple - it's not (topologically) a direct product, but rather a kind of twisted version of a direct product, called a fiber bundle. For illustration, while a band (a cylinder) is a direct product $S^1\times I$ (where $I=[0,1]$), a Mobius band $M$ (a twisted band) is a fiber bundle $I\to M\to S^1$. For $\mathrm{U}(n)$, if we have it act on the unit sphere $S^{2n-1}\subseteq\mathbb{C}^n$, the action is transitive with stabilizer $\mathrm{U}(n-1)$ (of the last standard coordinate basis vector $e_n$) so the orbit-stabilizer theorem says $\mathrm{U}(n)$ is a fiber bundle
$$ \mathrm{U}(n-1)\to\mathrm{U}(n)\to S^{2n-1}. $$
As $\mathrm{U}(1)=S^1$, this means $\mathrm{U}(n)$ is an extremely twisted version of $S^1\times S^3\times\cdots\times S^{2n-1}$. For $n=2$ though, it actually is a knit product $\mathrm{U}(2)=\mathrm{SU}(2)\mathrm{U}(1)$, where we embed $\mathrm{U}(1)\hookrightarrow\mathrm{U}(2)$ in the upper left or lower right corner (with $1$ in the other), and $\mathrm{SU}(2)=S^3$, so $\mathrm{U}(2)=S^3\times S^1$ is a bona fide topological direct product, which means $\mathrm{GL}_2\mathbb{C}\simeq S^3\times S^1\times\mathbb{R}^4$ topologically
