Fundamental group of $Spin^c(V)$ In picture below, Spin(V) is Spin group. In the  last paragraph , it wants to show $\pi_1(Spin^c(V))=\pi_1(S^1)$. But I almost don't know it.
First, where is the $S^1$ in ? For 'a homotopically nontrivial loop $\gamma$ in $S^1$', if the $S^1$ is $S^1$ of  $SO(V)\times S^1$, how to induces a loop in $Spin^c(V)$ ?   $Spin^c(V)$ is double covering of $SO(V)\times S^1$, then, there will induce two loop in $Spin^c(V)$. 
Second, why $\pi_1(Spin^c(V))$ contains $\pi_1(S^1)$ ? Seemly, it is because any loop in $Spin^c(V)$ will induces loop in $S^1$ of $SO(V)\times S^1$. But $SO(V)$ is not simply connect, why the loop will not in $SO(V)$ ?
Third,how to compose 2.4.14? I don't know the mean of compose.
Forth, how to know kernel is identified with $Spin(V) $ by 2.4.13 ?
In fact, I really know nothing about the last paragraph.I very thanks for detail explain of it.  
Picture below is from 75 page of Jost's Riemannian geometry and geometric analysis.



 A: From your other question, we have a map $\mathrm{Spin}(V)\times S^1\to \mathrm{Spin}^c(V)$ with kernel $\pm(1,1)$, which means we can write $\mathrm{Spin}^c(V)\cong \mathrm{Spin}(V)\times_{\mathbb{Z}_2}S^1$ where $(a,z)\sim(-a,-z)$, or equivalently where we have $(-a,z)\sim(a,-z)$.
At this point we can speak of $\mathrm{Spin}^c(V)$ and $\mathrm{Spin}(V)\times_{\mathbb{Z}_2}S^1$ interchangeably.
Next, there is a map $\mathrm{Spin}^c(V)\to \mathrm{SO}(V)\times S^1$ given by $(a,z)\mapsto (\overline{a},z^2)$, where by $\overline{a}$ I mean the image of $a$ under the usual double covering $\mathrm{Spin}(V)\to\mathrm{SO}(V)$. The kernel of this map is $(\pm1,\pm1)$ but $(-1,-1)\sim(1,1)$ and $(1,-1)\sim(-1,1)$ so the kernel is $\mathbb{Z}_2$, and this map can be called a double covering.
By picking the particular element $1\in\mathrm{Spin}(V)$, we can map $S^1\to \mathrm{Spin}^c(V)$ via $z\mapsto (1,z)$. This is injective, i.e. $(1,z_1)\sim(1,z_2)\Rightarrow z_1=z_2$. Thus, any loop in $S^1$ (i.e. based map $\gamma:S^1\to S^1$) can be turned into a loop in $\mathrm{Spin}^c(V)$, namely the composition $S^1\xrightarrow{\gamma}S^1\to\mathrm{Spin}^c(V)$.
(Recall the composition of two functions $g:X\to Y$ and $f:Y\to Z$ is $f\circ g:X\to Z$.)
We can compose $\mathrm{Spin}^c(V)\to\mathrm{SO}(V)\times S^1$ with $\mathrm{SO}(V)\times S^1\to S^1$ (the projection map) to obtain a map $\mathrm{Spin}^c(V)\to S^1$. We may then "chase" an element through the following:
$$ \begin{array}{ccccccc}
S^1 & \longrightarrow & \mathrm{Spin}^c(V) & \longrightarrow & \mathrm{SO}(V)\times S^1 & \longrightarrow & S^1 \\
z & \mapsto & (1,z) & \mapsto & (\mathrm{id},z^2) & \mapsto & z^2
\end{array} $$
Thus, if we start with a nontrivial loop $\gamma$ in $S^1$, then put it inside $\mathrm{Spin}^c(V)$, then project it back to $S^1$ again, we get $2\gamma$ which is again nontrivial. If the image of $\gamma$ inside $\mathrm{Spin}^c(V)$ were trivial then it would project onto a trivial loop in $S^1$, but it doesn't, so it is nontrivial within $\mathrm{Spin}^c(V)$.
We've seen that loops in $S^1$ induce loops in $\mathrm{Spin}^c(V)$. Any homotopy between loops in $S^1$ can be performed in $\mathrm{Spin}^c(V)$ as well, so really homotopy classes of loops in $S^1$ induce homotopy classes of loops in $\mathrm{Spin}^c(V)$, i.e. we have a map $\pi_1(S)\to\pi_1(\mathrm{Spin}^c(V))$. Since composition of loops in $S^1$ corresponds to composition of loops in $\mathrm{Spin}^c(V)$, this is in fact a group homomorphism, and we've seen its kernel is the trivial loop only, so it is an embeddding and we may identify $\pi_1(S^1)$ with a subgroup of $\pi_1(\mathrm{Spin}^c(V))$.
Suppose we had a loop in $\mathrm{Spin}^c(V)$ that didn't come from $S^1$, i.e. is not in this copy of $\pi_1(S^1)$. The text seems to be assuming that, without loss of generality, this loop's projected image in $S^1$ is trivial, but I don't see why.
At any rate, suppose we have a loop in $\mathrm{Spin}^c(V)$ which maps to the trivial loop in $S^1$. Then the loop is (up to homotopy) entirely contained within the kernel of $\mathrm{Spin}^c(V)\to S^1$, which is all $(a,z)$ with $z^2=1$, i.e. $(a,\pm1)$, but $(a,-1)\sim(-a,1)$ so kernel is exactly the set of all $(a,1)$ with $a\in\mathrm{Spin}(V)$. This is an embedded copy of $\mathrm{Spin}(V)$ within $\mathrm{Spin}^c(V)$, which is simply connected if $\dim V>2$, so this loop is trivial.
