Metaplectic group is the unique double covering of symplectic group

Question

The proof that $Mp(2n,\mathbb{R})$ is the unique connected double cover of $Sp(2n,\mathbb{R})$ uses the fact that the fundamental group of the latter is infinite cyclic (the integers). I have not found any proofs of both fundamental facts about this statement:

1. The fundamental group of $Sp(2n,\mathbb{R})$ is the integers;
2. If a group has an infinite cyclic fundamental group, then it has a unique connected double cover;

Since those points are being addressed in the proof of a single proposition, I'm asking them as a single question.

Answer 1

The following statements are all proved in the books

• [1] Symplectic Geometry and Quantum Mechanics, by Maurice de Gosson;
• [2] Introduction to Symplectic Topology, by Dusa McDuff and Dietmar Salamon;

Proposition 1: The quotient $Sp(2n, \mathbb{R})/U(n)$ is contractible.

Proof: Done by constructing an explicit retraction of $Sp(2n, \mathbb{R})$ onto $U(n)$. Check Corollary 2.24 of [1].

Proposition 2: $\pi_1(U(n)) \simeq \mathbb{Z}$.

Proof: Uses an exact homotopy sequence and the fact that $SU(n)$ is simply connected. Check Proposition 2.23 of [2].

Theorem 1: $\pi_1(Sp(2n, \mathbb{R})) \simeq \mathbb{Z}$.

Proof: By Proposition 2, $\pi_1(U(n)) \simeq \mathbb{Z}$. By Proposition 1, $U(n)$ and $Sp(2n, \mathbb{R})$ are homotopic.

Theorem 2: There exists and unique double cover of $Sp(2n, \mathbb{R})$.

Proof: $\pi_1(Sp(2n, \mathbb{R})) \simeq \mathbb{Z}$, which is isomorphic to the group of covering transformations of $Sp(2n, \mathbb{R})$. Since $\mathbb{Z}$ is infinite cyclic and $Sp(2n, \mathbb{R})$ is connected, there is a 1-1 correspondence between subgroups of $\mathbb{Z}$ and subgroups of the covering transformations group. Since $\mathbb{Z}$ has a single 2-fold subgroup, so does the group of covering transformations of $Sp(2n, \mathbb{R})$. (This is a mix between @LordSharktheUnknown's comments and some proofs I found on Allen Hatcher's book Algebraic Topology.)