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?
 A: Rather than using the ill-suited "one-to-one" qualification, which strictly speaking does not apply in this context, one can state that the parameter $\theta$ is identifiable, i.e., that $\theta_1\ne\theta_2$ implies that the corresponding distributions indexed by  $\theta_1$ and $\theta_2$ differ.
The exact result is that the parameter $\theta$ is identifiable if the statistics $F_1$, ..., $F_n$ are linearly independent and the parameter space $\Theta$ is an open set.
The constraint on $\Theta$ being an open set excludes the artificial addition of a statistic $F_{n+1}$ attached with a parameter $\theta^{n+1}$ taking a single possible value, eg $\theta^{n+1}=1$.
To quote from Geyer:

Theorem 1. $\ $ Every exponential family has a minimal representation, and a representation is minimal if and only if neither
of the following conditions hold.
(i) There exists a hyperplane that contains the natural statistic
vector with probability one.
(ii) There exists a hyperplane that contains the natural parameter
space.

and

Theorem 5. A full exponential family is identifiable if and only if condition (i) of Theorem 1 does not hold. Moreover, if $θ$ and $ψ$ are
distinct parameter values corresponding to the same distribution then
$sθ + (1 − s)ψ$ is contained in the full natural parameter space and
corresponds to the same distribution for all real s.
So, if one uses a minimal representation, then the family is
identifiable.

A: 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. $3-4.$
A: 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.
