Let $X$ be a Hausdorff locally compact topological space, $s:X\times X\to \mathbb{R}$ be a continuous map such that $s(x,x)=0$ for all $x\in X$. Set $\triangle=\{(x,y): x=y, x,y\in X\}$ and $A=\{(x,y): s(x,y)=0\}$, so $\triangle\subset A$. If for each fixed $x\in X$, $s(x,\cdot)$ is a local homeomorphism near $y=x$ in the second variable, show that the set $Y=A\backslash \triangle$ is closed in $X\times X$.
(I'm not certain the condition in the problem is sufficient, the real problem I met is that $X$ is a complex manifold and $\mathbb{R}$ being replaced by another complex manifold, $s$ is a holomorphic map and it's a local biholomorphism near the diagonal in the second variable. But I believe it's really a topological problem..)
The statement is not valid. Let $X=\mathbb{R}$ and let $s(x,y)=y(x-y)(x+y)=x^2y-y^3$ for all $x,y\in\mathbb{R}$. Clearly, $s$ is continuous. For a fixed $x\in\mathbb{R}$, the derivative of $s(x,y)$ with respect to $y$ is $-3y^2+x^2$. For $x\neq 0$, this derivative is nonzero at $y=x$, so $s(x,\cdot)$ is a local homeomorphism at $y=x$. For $x=0$, $s(0,y)=-y^3$, so again $s(0,\cdot)$ is a homeomorphism. However, the set $A\setminus\Delta=\{(x,y)\in\mathbb{R}^2\!:y=-x\ \wedge\ x\neq 0\}$ is not closed in $\mathbb{R}^2$.