2024-12-08

892: Composition of Map After Preimage Is Identical if Map Is Surjective w.r.t. Argument Subset

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description/proof of that composition of map after preimage is identical if map is surjective w.r.t. argument subset

Topics


About: set

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

Main Body

1: Structured Description


Here is the rules of Structured Description.

Entities:
S1′: ∈{ the sets }
S2′: ∈{ the sets }
f: :S1′→S2′
S2: ⊆S2′
//

Statements:
∀p∈S2(p∈f(S1′))
⟹
f∘f−1(S2)=S2
//

When f is a surjection, the condition is satisfied for any subset of S2′.

2: Natural Language Description


For any sets, S1′,S2′, any map, f:S1′→S2′, and any subset, S2⊆S2′, if f is surjective with respect to S2, which means that any point on S2 is in the range of f, f∘f−1(S2)=S2.

When f is a surjection, f is surjective with respect to any subset of S2′.

3: Proof


Whole Strategy: Step 1: see that f∘f−1(S2)⊆S2 in general; Step 2: suppose that ∀p∈S2(p∈f(S1′)), and see that S2⊆f∘f−1(S2).

Step 1:

f∘f−1(S2)⊆S2, by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.

So, it is about S2⊆f∘f−1(S2).

Step 2:

Let us see that S2⊆f∘f−1(S2).

Let us suppose that ∀p∈S2(p∈f(S1′)).

For any p∈S2, there is a p′∈f−1(p)⊆f−1(S2). p=f(p′)∈f∘f−1(p)⊆f∘f−1(S2).

References
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