Is $(x^2y,xy^2)$ log smooth?

Question

Consider the map $$f:\mathbb C^2\to\mathbb C^2$$ $$(x,y)\mapsto(x^2y,xy^2)$$ We can view $f$ as induced by the map of monoids $g:\mathbb Z^2_{\geq 0}\to\mathbb Z^2_{\geq 0}$ given by the matrix $(\begin{smallmatrix}2&1\cr 1&2\end{smallmatrix})$. Thus $f$ is a map of log schemes.

Is $f$ log smooth?

If I understand Proposition 6.1 from https://arxiv.org/abs/alg-geom/9406004 correctly, the answer is supposed to be yes, since $\ker g$ and $(\operatorname{coker} g)_{\mathrm{tors}}$ are both finite.

On the other hand, log smooth maps are supposed to be flat, and flat maps are open. Thus if $f$ were log smooth it would have to be open. But it is not open: $(0,0)$ is in the image of $f$, but $(0,a)$ is not whenever $a\ne 0$.

Answer 1

Warning: I'm not an expert on logarithmic geometry.

According to Proposition 4.1.2. of Ogus' Lectures on Logarithmic Algebraic Geometry, a log smooth morphism (of fine log schemes) is log flat. Log flatness is defined in Definition 4.1.1. Evidently, log flatness does not imply classical flatness. However, the same proposition says that log flatness does imply classical flatness (and is in fact equivalent to it!) if the morphism is strict.

We can confirm that $f$ is not strict using Proposition 1.2.4: Let $\theta$ be the monoid homomorphism $\Bbb N^2\to \Bbb N^2$ given by $\begin{bmatrix}2 &1\\1 &2\end{bmatrix}$. This map is already logarithmic since neither side has any units other than 0. However, $\theta$ is not an isomorphism and is therefore not strict. Hence, $f$ is not strict.

Answer 2

It is smooth. For a log map to have good "fiber bundle properties", one needs more than smoothness. Rather, one needs the relevant maps of monoids to be exact in the sense of Kato. A reference is Nakayama--Ogus "Relative rounding in toric and logarithmic geometry" Mathscinet Journal.