Spin$(n)$ is connected for $n\geq 2$

Question

I am reading the proof that Spin$(n)$ is connected for $n\geq 2$ from the book Dirac Operators in Riemannian Geometry.

I want to understand why it's sufficient to find a path between $-1$ and $1$, given the conditions. I don't think we are supposed to use any topological information about $SO(n)$.

Answer 1

Some topological information about $SO(n)$ is necessary, namely that it is a connected manifold.

Lemma 1. Let $X,Y$ be two topological spaces and $f:X\to Y$ an open surjective map. If $Y$ is connected and any two points $x,y\in X$ such that $f(x)=f(y)$ belong to a connected subset of $X$ then $X$ is connected.

Proof. Assume that $X=U\cup V$ for some nonempty and open subsets $U,V\subseteq X$. We will show that their intersection is nonempty. We have $Y=f(U)\cup f(V)$ and since $f$ is open and $Y$ is connected then $f(U)\cap f(V)\neq\emptyset$. Meaning there is $x\in U$ and $y\in V$ such that $f(x)=f(y)$. By our assumption $x,y\in C$ for some connected subset $C\subseteq X$. But then $C\cap U\neq\emptyset\neq C\cap V$ and thus $(C\cap U)\cap(C\cap V)\neq\emptyset$ since $C$ is connected. Meaning $U\cap V\neq\emptyset$. $\Box$

Lemma 2. Let $G,H$ be topological groups such that both are locally compact Hausdorff and $G$ is $\sigma$-compact. Furthermore let $f:G\to H$ be a surjective continuous group homomorphism. If $H$ is connected and any two points in $\ker f$ belong to a connected subset of $G$ then $G$ is connected.

Proof. In order to apply our lemma 1, first of all we need to know that $f$ is open. This is guaranteed by the open mapping theorem for topological groups under our assumptions on $G$ and $H$. Now we can apply lemma 1, because if the lemma's assumption is satisfied for $\ker f$ then it is satisfied for any pair $x,y$ such that $f(x)=f(y)$. $\Box$

Now we can easily apply lemma 2 to your case, since Lie groups (or at least groups we are dealing with) are locally compact and $\sigma$-compact.