A function $f:\mathbb R^{1,1}\to \mathbb R^{1,1}$ is "pseudocontinuous" if and only if: $$f(x,y)\equiv(g(x),h(y)),$$ or $$f(x,y)\equiv(h(y),g(x)),$$ for any pair of continuous functions $f,g:\mathbb R\to\mathbb R$. Why? I list the sequence of lemmas which one should prove: **Lemma 1**: We can redefine the pseudometric to $(x,y)\mapsto |xy|$ instead of $(x,y)\mapsto |x^2 - y^2|$. We now switch to this pseudometric. **Lemma 2**: A "cross" (that is, the union of a line parallel to the $x$ axis with a line parallel to the $y$ axis) necessarily gets mapped to another cross by any pseudocontinuous function $f$. **Lemma 3**: A horizontal line or vertical line gets mapped to a horizontal or vertical line by pseudocontinuous $f$. **Lemma 4**: For fixed $x$, the function $y \mapsto f(x,y)$ is continuous. The same is true for $x \mapsto f(x,y)$ for fixed $y$. **Lemma 5**: The function $f$ is continuous everywhere. **Theorem**: The above statement.