Why is $\theta\mapsto p_{\theta}$ one-to-one $\iff$ $(n+1)$ functions $\{F_1,\ldots,F_n,1\}$ are linearly independent? If an n-dimensional model $S=\{p_{\theta}| \theta\in\Theta\}$ can be expressed in terms of the functions $\{C,F_1,\ldots,F_n\}$ on a sample space $X$ and a function $\psi\in \Theta$(parameter space) as
$$p(x;\theta)=\exp\left[C(x)+\sum_{i=1}^{n}\theta^{i}F_{i}(x)-\psi(\theta)\right]$$ then we say that $S$ is an exponential family.
My question is , it is known that $\theta\mapsto p_{\theta}$ is one-to-one if and only if the $(n+1)$ functions $\{F_1,\ldots,F_n,1\}$ are linearly independent. But I am not getting this line. Can someone explain it or give some hint?

To see this result in a simple way, it is useful to note that $\psi$ is merely a scaling constant that is fully determined by the other functions (to make the density integrate to one).  You can write the density for the exponential family form without this by using the proportionality requirement:
$$p_\theta(x) \overset{x}{\propto} \exp \bigg( C(x) + \sum_{i=1}^n \theta_i F_i(x) \bigg).$$
Now, if the vectors of interest are linearly dependent then there is a vector $\alpha = (\alpha_0,...,\alpha_n) \neq \mathbf{0}$ such that:
$$\alpha_0 + \sum_{i=1}^{n} \alpha_i F_{i}(x) = 0
\quad \quad \quad \quad \quad 
\text{for all } x \in \mathscr{X}.$$
Consider a parameter value $\theta$ in the interior of the parameter space and another parameter value $\theta^* \equiv \theta + \alpha \neq \theta$ that is also in the parameter space.$^\dagger$  Then for all $x \in \mathscr{X}$ we have:
$$\begin{align}
C(x) + \sum_{i=1}^{n} \theta^*_{i} F_{i}(x) 
&= - \alpha_0 + C(x) + \alpha_0 + \sum_{i=1}^{n} \theta^*_{i} F_{i}(x) \\[6pt]
&= - \alpha_0 + C(x) + \alpha_0 + \sum_{i=1}^{n}(\theta_{i} + \alpha_i)F_{i}(x) \\[6pt]
&= - \alpha_0 + C(x) + \alpha_0 + \sum_{i=1}^{n}\alpha_i F_{i}(x) + \sum_{i=1}^{n} \theta_{i} F_{i}(x) \\[6pt]
&= - \alpha_0 + C(x) + \sum_{i=1}^{n}\theta_{i}F_{i}(x), \\[6pt]
\end{align}$$
which implies that $p_{\theta^{*}} \overset{x}{\propto} p_\theta$, which implies that $p_{\theta^{*}} = p_\theta$.  Since $\theta^* \neq \theta$ this means that the function $\theta \mapsto p_\theta$ is not invertible (and therefore not one-to-one).

$^\dagger$ As Xi'an points out, the non-identifiability you are considering also requires that we let $\Theta$ be an open set.  If this is the case then there is a point $\theta \in \text{int} \Theta$ and a vector $\alpha$ (satisfying the previous linear equation) that is sufficiently small in magnitude that $\theta+\alpha \in \Theta$.  For this part, observe that any scalar multiple of $\alpha$ satisfies the linear equation given, so we can make the magnitude of this vector arbitrarily close to zero.
