$\overline{\theta}$ the maximum likelihood estimator of $\theta \implies$? I can't understand how the following statement holds without any extra conditions on the function $g$:
$\overline{\lambda}$ the maximum likelihood estimator of parameter $\lambda \implies g(\overline{\lambda})$ the maximum likelihood estimator of $g(\lambda)$
Intuitively it would seem that we would need $g$ to be monotionic for this to be true. How can it be shown that this statement holds for an arbitrary function $g$?

Suppose the likelihood function is $\alpha\mapsto L(\alpha)$.
Let $\beta=g(\alpha)$. I will assume $g$ is one-to-one.
The likelihood as a function of $\beta$ is not the same function as the likelihood as a function of $\alpha$.
The value of $\alpha$ determines the probability distribution of the data, so that $L(\alpha_0) = \Pr(X=x\mid\alpha=\alpha_0)$, where $x$ is the data actually observed (or if $X$ is a continuous variable, then $L(\alpha_0)=f(x\mid\alpha=\alpha_0)$, where $f$ is the probability density function).
I will give to the likelihood as a function of $\beta$ the name $K$.  Thus
$$K(\beta_0) = \Pr(X=x\mid \beta=\beta_0) = \Pr(X=x\mid g(\alpha)=\beta_0).$$
If $L(\alpha_1)<L(\alpha_2)$ then $\Pr(X=x\mid\alpha=\alpha_1)<\Pr(X=x\mid\alpha=\alpha_2)$, so $\Pr(X=x\mid g(\alpha)=g(\alpha_1))<\Pr(X=x\mid g(\alpha)=g(\alpha_2))$, and so $K(g(\alpha_1))<K(g(\alpha_2))$.
Hence if $\alpha_2$ is a value for which $L(\alpha_1)<L(\alpha_2)$ for EVERY value of $\alpha_1$, then $g(\alpha_2)$ is a value for which $K(g(\alpha_1))<K(g(\alpha_2))$ for EVERY value of $g(\alpha_1)$.
