# Equalizer of local analytic isomorphisms

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?

• Welcome to MathOverflow! Your question can be undestood either as "Is the equalizer (in the category of analytic spaces) smooth?", or as "Is there an equalizer in the category of smooth analytic spaces?" Your comments seem to point to the first interpretation, but I am not quite sure. – Laurent Moret-Bailly Jan 18 at 20:11
• @LaurentMoret-Bailly Thank you! I’m interested in smooth equalizers. My comments attempted to give a construction and show where it may break down: I was first constructing the underlying topological space, then hopefully a complex analytic space equalizer, and only later I was hoping to show it’s smooth, but it feels now that a smooth equalizer should not exist in general – John P. Jan 19 at 2:44

Take, say, $$W=\mathbb{C}^n$$, $$U=$$ some neighborhood of 0 in $$\mathbb{C}^n$$, $$a=$$ the inclusion. Now fix some analytic $$h:U\to W$$ such that $$h(0)=0$$ and $$Z:=h^{-1}(0)$$ is not smooth at $$0$$. For small engouh $$\varepsilon>0$$, $$b:=a+\varepsilon h$$ is a local isomorphism (possibly after shrinking $$U$$), but the equalizer of $$(a,b)$$ is $$Z$$.