Let $a,b : V\to W$ be two morphisms of **smooth** complex analytic spaces.


Assume $a$ and $b$ are local analytic isomorphisms.



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> * Does the equalizer $U$ of $a,b$ exist as a smooth complex analytic space?
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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?



Equalizer of local analytic isomorphisms
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 $Z$.


