Let $a,b : V\to W$ be two morphisms of smooth complex analytic spaces.
Assume $a$ and $b$ are local analytic isomorphisms.
- Does the equalizer $U$ of $a,b$ exist as a smooth complex analytic space?
It feels the answer should be “no”, even though $a$ and $b$ are local analytic isomorphisms.
As a topological space $U$ is the fiber product over $W\times W$ of $(a,b) : V\to W\times W$ and the diagonal $\Delta : W\to W\times W$ and neither of these two maps is a submersion, since the dimension of the tangent space at any point of $W\times W$ is going to be twice the dimension of the tangent space at any point of $V$ and at any point of $W$.
There’s no reason to expect that $\Delta$ and $(a,b)$ should be transverse, and so $U$ may not even exist as a differentiable manifold.
However, is there something special about complex analytic spaces that ensures the existence of $U$ as a smooth complex analytic space?