Following Qing Liu's book *Algebraic geometry and arithmetic curves*In section 6.3.2, we say that a map. $ f: X rightarrow Y $ of varieties on some field $ k $ is *local full intersection in $ x in X $* if there is a neighborhood $ U $ of $ x $ and a chain of morphisms. $ U overset {i} { hookrightarrow} Z overset {g} { rightarrow} Y $, where $ g circ i = f | _U $, $ Z $ it's a scheme about $ k $, $ i $ it is a regular inlay and $ g $ It is a smooth map. We say $ f $ is *local full intersection* (l.c.i for short) if it is l.c.i in each $ x in X $.

Now, let me come to my problem. Leave $ X $ Being a smooth complex projective variety, $ Y $ be another complex projective variety (not necessarily soft), and leave $ f: X rightarrow Y $ Be a map. Leave $ Z subset Y $ Be a soft closed sub-variety. Leave $ f: X rightarrow Y $ Be a map with the property that. $ f $ it's an isomorphism about $ Y setminus Z $, plus $ f $ is a $ mathbb {P} ^ n $– package about $ Z $ for some $ n $.

I believe that $ f $ it is a morphism of l.c.i, for which I have a discussion, but I am not completely sure if it is correct. Can someone please stop by my discussion and let me know if it makes sense or not?

**Claim:** $ f $ It is a morphism of l.c.i.

*Test:* According to the Stacks project {https://stacks.math.columbia.edu/tag/069N}, the morphism of lci is fpqc-local at the base, and since any zariski cover is also a fpqc cover, enough for Show that $ f $ is l.c.i on a related deck of $ Y $.

We consider two cases:

1) yes $ y in Z $, choose a related neighborhood $ U $ of $ and $ such that $ f ^ {- 1} (U) cong U times mathbb {P} ^ n $ Y $ f | _ {f ^ {- 1} (U)} $ is the usual projection on the 1st component (remember that $ f $ It is a projective package $ Z $). As $ mathbb {P} ^ n $ has finished l.c.i. $ mathbb {C} $Y $ U $ it's flat on $ mathbb {C} $, we have that $ U times mathbb {P} ^ n rightarrow U $ it is also l.c.i due to the fact that a flat base change of a map l.c.i is a map l.c.i (see the Battery Project, https://stacks.math.columbia.edu/tag/069I). From there we conclude that $ f ^ {- 1} (U) rightarrow U $ It is the map of l.c.i.

2) yes $ y in Y setminus Z =: V $, then cover $ V $ by affinity it opens, and since then $ f $ is isomorphism about $ V $, clearly $ f $ is a map of l.c.i on each of the open affinities.

From there we conclude that $ f $ It is the morphism of l.c.i. *(shown)*

Can someone please tell me if this is correct? Thanks in advance.