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?

Let $\mathcal P:= \{P_\theta:\theta\in \Theta\} $ be an exponential family. Then for $P_\theta\in \mathcal P,$
$$\frac{\mathrm dP_\theta}{\mathrm d\mu}(x) = b(x)a(\theta)\exp\left[\alpha(\theta)\cdot t(x)\right];\tag 1 $$  the representation is minimal if for $\alpha := \langle \alpha_i\rangle_{i\in\{1, \ldots, k\}}, ~t := \langle t_i\rangle_{i\in\{1, \ldots, k\}},~~\alpha_i$s are affinely independent and so are $t_i$s.
Theorem $1.$: If an exponential family $\mathcal P$ has a minimal representation, then the parameterization is one-one in that $P_\theta = P_{\phi}; ~~\theta,~\phi\in \Psi\wedge\theta\ne \phi.$
The proof is straightforward simply by noticing
$$\langle \theta, F\rangle  = \langle \phi, F\rangle + c~~~~\textrm{a.s.}, \tag 2$$
where $c := \ln \psi(\phi) - \ln \psi(\theta). $
Rest follows from the definition of minimal representation.
$\blacksquare$
In fact, as noted in $\rm [I],$

[...] the mapping $\omega \to P_\omega$ is one-to-one if and only if $\omega\to\alpha(\omega)$ is one-to-one.


References:
$\rm [I]$ Information and Exponential Families in Statistical Theory, O. Barndorff-Nielsen, John Wiley & Sons., $2014,$ sec. $8.1,$ p. $111, ~113.$
$\rm [II]$ Exponential Families, R. Dudley, $18.466$ Notes, Feb. $26, ~2013,$ pp. $3-4.$
