If the dimension of a sufficient statistic $T(X)$ equals the dimension of parameter space, $T(X)$ is minimal sufficient? I came cross an interesting comment saying

If the dimension of a sufficient statistic $T(X)$ is the same as that
of the parameter space, then $T(X)$ is minimal sufficient.

Is this is true? I examined some examples e.g. full-rank exponential families, curved exponential families e.g. $N(\theta, \theta^2)$, and some uniform distributions, e.g. $U(-\theta,\theta)$, $U(0,\theta)$, $U(\theta,\theta+a)$ and found their sufficient statistic has the same dimension with the parameter space and indeed they are minimal sufficient.
Is this quoted statement correct? Can you please provide a reference of a proof, or a counter example?
 A: The construction by Whuber in the comments gives an example to create a sufficient statistic that is not minimal sufficient.
The trick of that counterexample is to design a sufficient statistic that has a larger space, but still the same dimension. For a normal distribution with fixed $\mu$ and parameterized by only $\sigma$, the parameter is the set of positive numbers $\sigma \subset \mathbb{R_{>0}}$ and the minimal sufficient statistic has the same space $s \subset \mathbb{R_{>0}}$. This leaves room for the creation of a statistic $s^\prime \subset \mathbb{R}$ by adding the negative numbers (in the example by Whuber this is done my multiplying with the sign of a variable $x_1$), which has a non-injective surjective relationship with the minimal sufficient statistic $s$ such that the inverse function does not exist.

A: The quoted statement is incorrect. The post Minimal sufficient statistic whose dimension is less than dimension of parameter  has one example, @whuber another one in a comment.
This: Minimum dimension of sufficient statistics is also relevant.
