Let $g$ denote genus and $n$ denote number of punctures. According to Kra's construction in Pseudo-Anosov theory for surfaces, if $S$ is an orientable surface such that $3g+n>3$ and $\gamma \in \pi_1(S,p)$ is a closed curve on $S$, then mapping class $P(\gamma)$ (where $P$ is the point-push map) is pseudo-Anosov iff $\gamma$ fills $S$ (for reference, see theorem 1.1 at http://www.math.uiuc.edu/~dowdall/point_pushing_pseudo-Anosovs.pdf).
Now, I want to show the inequality is strict. Let $S$ be a Torus with no punctures. Thus $S$ has $g=1$ and $n=0$, so the theorem does not apply. However, it is still true that $P(\gamma)$ is pseudo-Anosov implies $\gamma$ fills $S$. Thus, according to the theorem, there must be a closed curve $\gamma$ filling $S$ such that $P(\gamma)$ is NOT pseudo-Anosov. I would like to find such a $\gamma$.
Thus, I want to find a closed curve $\gamma$ on $S$ such that $P(\gamma)$ is Not pseudo-Anosov. I could not find such a curve. Can you? Or am I missing something?
I found this old question, which has an easy answer.
Every closed curve $\gamma$ on the torus which is not homotopic to a constant is homotopic to a multiple of some simple closed curve. The point pushing map around any such $\gamma$ is isotopic rel $p$ to the identity map on the torus. Thus, $P(\gamma)$ is the identity mapping class for every $\gamma$.
So yes, it is vacuously true that if $P(\gamma)$ is pseudo-Anosov then $\gamma$ fills $S$. In fact, $P(\gamma)$ is never pseudo-Anosov, and $\gamma$ never fills $S$.