范畴论中filter操作是否为态射？属于何种类型？

Great question! Let's unpack how filter fits into category theory, using your Scala examples as context.

Short Answer #

Yes, filter is considered a morphism, and in the context of Scala's type system (and the broader category of sets/types and functions), it's an endomorphism—a morphism that maps an object to itself.

Breakdown #

First, a quick refresher: in category theory, a morphism is simply an arrow between two objects that adheres to the category's composition rules. For filter:

• The "objects" here are Scala types like Seq[Int]. Every filter call takes a Seq[Int] and returns another Seq[Int]—so the source and target object are the same, making it an endomorphism.
• It respects composition rules: if you have two predicates p and q, running filter(p) followed by filter(q) is equivalent to filtering once with p && q. This aligns perfectly with how function composition works in the category of Scala functions.

Your Scala Examples in Context #

Let's look at your code to make this tangible:

val myNums: Seq[Int] = Seq(-1, 3, -4, 2)
myNums.filter(_ > 0)       // Seq[Int] = List(3, 2) → subset, same type
myNums.filter(_ > -99)     // Seq[Int] = List(-1, 3, -4, 2) → identical to original
myNums.filter(_ > 99)      // Seq[Int] = List() → empty sequence, same type

No matter the predicate, filter always outputs a Seq[Int]—it never changes the type, just selects elements within the same "object" (type) structure. Even the empty sequence is a valid member of the Seq[Int] object, so the morphism stays within the same object's bounds.

A Narrower Set Theory Perspective #

If we zoom into the category of sets (where objects are sets, morphisms are functions between sets), filter(p) corresponds to a function from the set of all integer sequences to itself. Again, this is an endomorphism—mapping elements of the set to other elements of the same set.

内容的提问来源于stack exchange，提问作者Polymerase