Double cover of $GL(n,\mathbb{C})$. Is there any reference or simple proof to show that the group $\frac{\mathbb{C}\times SL(n,\mathbb{C})}{2\mathbb{Z}}$ is double cover of $GL(n,\mathbb{C})$.
 A: I'm likely duplicating the proof in Igor's reference.  The idea is that there is a homomorphism $GL_n(\mathbb C) \to \mathbb{C}^* \equiv \mathbb{C} \setminus \{0\}$ by taking the determinant.  Since $\mathbb{C}^*$ is the punctured plane, it has a canonical double cover and there is an induced canonical double cover of $GL_n(\mathbb C)$.
Moreover, since the determinant's kernel is $SL_n(\mathbb C)$ there is a nice way of expressing this covering space as $SL_n(\mathbb C) \times_{\mathbb Z_2} \mathbb{C}^*$.  The map from $SL_n(\mathbb C)$ to the cover is given by inclusion (lifting property).  One can recognise $\mathbb{C}^*$ in $GL_n(\mathbb C)$ as the diagonal matrices.  In that way $\mathbb{C}^*$ lifts isomorphically to a subgroup of the double cover.  The two groups $SL_n(\mathbb C)$ and $\mathbb{C}^*$ meet in two places in this lift, that's the $\mathbb Z_2$ intersection.
A: A proof is given here (section 9.6): http://books.google.com/books?id=nS_5yegroSoC&pg=PA292&lpg=PA292&dq=metalinear+group&source=bl&ots=vgW39hgnSu&sig=xFMxJWOYB9b3iz_c0rPUaS-JfPk&hl=en&sa=X&ei=PfLGUqa9OJK_sQTx3oCADg&ved=0CE8Q6AEwBQ#v=onepage&q=metalinear%20group&f=false
For posterity, the reference is:
A User's Guide to Algebraic Topology
 By C.T. Dodson, P.E. Parker
