How to show $Spin^c(V)$ is isomorphic to $Spin(V)\times_{Z_2} S^1$? In first picture below, Spin(V) is the Spin group,Cl(V) is Clifford algebra.
First, how to get 2.4.13 is surjective from the commute of $S^1$ and Spin(V) ?
Second,why Spin(V) $ \cap S^1$ is $\{1,-1\}$ ?
All the pictures is from about 75 page of  Jost's Riemannian geometry and geometric analysis. 



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 A: (1) Suppose a group $G$ is generated by a subset $S$. That means there is no proper subgroup of $G$ which contains $S$, and it also means every element of $G$ may be expressed as a "word" in the "letters" from $S$, i.e. as $g=s_1^{\epsilon_1}s_2^{\epsilon_2}\cdots s_k^{\epsilon_k}$. It is a basic group theory exercise to prove this.
If $G$ is generated by two commuting subgroups $H$ and $K$, i.e. if $G=\langle H\cup K\rangle$ and $[H,K]=1$ then there is a surjective group homomorphism $H\times K\to G$ given by $(h,k)\mapsto hk$. To prove it is a group homomorphism, use the fact that $hk=kh$ for all $h\in H,k\in K$. To see that it is surjective, observe every element of $G$ must be a "word" in the elements of $H$ amd $K$, but since elements of $H$ and $K$ commute we may slide all of the $H$ elements to the left and all of the $K$ elements to the right until we have a product of elements of $H$ times a product of elements of $K$, in other words an element of the form $hk$.
In the case of a topological group or a Lie group, we might want a closed subgroup generated by a subset $S$. Thus we would do the above construction to get the "abstract" group generated by $S$, and then take its closure to get the smallest closed subgroup containing $S$. In our case, the abstract subgroup we generate is already a closed subgroup. (It is the continuous image of a compact space, hence compact, so closed and bounded.)
(2) If $A\otimes B$ is a tensor product of $k$-algebras then there are embeddings $A\to A\otimes B$ and $B\to A\otimes B$ given by $a\mapsto a\otimes 1$ and $b\mapsto 1\otimes b$. The intersection of these two things is
$$ (A\otimes 1)\cap (1\otimes B)=k(1\otimes 1), $$
i.e. the scalar multiples of the identity. Indeed, if $\{a_1,\cdots,a_r\}$ and $\{b_1,\cdots,b_s\}$ are $k$-vector space bases for $A$ and $B$ with $a_1=1$ and $b_1=1$ then $\{a_i\otimes b_1\}$ is a vector space basis for $A\otimes 1$, $\{a_1\otimes b_j\}$ is a vector space basis for $1\otimes B$, and $\{a_i\otimes a_j\}$ is a vector space basis for $A\otimes B$.
Now we can treat it as a linear algebra problem. If $W$ is a vector space with basis $P$, and $Q,R\subseteq P$ are two subsets, and $U,V$ are subspaces of $W$ spanned by $Q,R$ respectively, then $U\cap V$ is spanned by $Q\cap R$.
