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?