How to visualize local complete intersection morphisms?

Question

As the question title asks for, how do others visualize local complete intersection morphisms? My experiment in asking people in real life didn't pan out, so I'm consulting the MO algebraic geometry community. Bonus points for pictures!

Answer 1

As Dan Petersen said the comments, an lci morphism $f:X\to Y$ is precisely one that factors Zariski-locally as $X\to Y\times \mathbf{A}^n \to Y$, where $X\to Y\times \mathbf{A}^n$ is a regular embedding (on affines, this means that you are looking at a closed subscheme defining by the vanishing of an ideal generated by a regular sequence) and $Y\times \mathbf{A}^n \to Y$ is the projection map. A regular embedding should be thought of as a closed subscheme which has a tubular neighborhood (every closed immersion of smooth schemes is a regular embedding), so an lci morphism is precisely a morphism for which it is reasonable to talk about the normal bundle.

This can be made more precise. Namely, an lci morphism $f:X\to Y$ can be characterized as a morphism such that the relative cotangent complex $L_{Y/X}$ is perfect, and has Tor-amplitude in $[-1,0]$. The latter statement means that for every quasicoherent sheaf $\mathscr{F}$ on $Y$, the derived tensor product $L_{Y/X}\otimes^\mathbf{L} \mathscr{F}$ has cohomology concentrated in degrees $-1$ and $0$. (If your definition of an lci morphism is as a morphism which is locally given by quotienting by a regular sequence, then one direction of the above statement is proved as Tag 08SL.) In any case, the conormal bundle $N_{Y/X}$ can now be defined as $L_{Y/X}[-1]$, so that the normal bundle of $f$ is $N_{Y/X}^\vee = L_{Y/X}^\vee [-1]$. Since $f$ is an lci morphism, this is actually a vector bundle over $Y$, so it makes sense to call this the normal bundle. This should make sense: if $f:Y = \mathrm{Spec}(B) \to X = \mathrm{Spec}(A)$ is a surjective morphism and $I = \ker(A\to B)$, then $H^1(L_{Y/X}) = I/I^2$.