2024-12-15

898: For Surjection, Preimages of Subsets Are Same iff Subsets Are Same

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description/proof of that for surjection, preimages of subsets are same iff subsets are same

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 surjection, the preimages of any subsets of the codomain are same iff the subsets are same.

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, ∈{ the surjections }
S2,1: ⊆S2
S2,2: ⊆S2
//

Statements:
f−1(S2,1)=f−1(S2,2)
⟺
S2,1=S2,2
//

2: Natural Language Description


For any sets, S1,S2, any surjection, f:S1→S2, and any subsets, S2,1,S2,2⊆S2, f−1(S2,1)=f−1(S2,2) if and only if S2,1=S2,2.

3: Proof


Whole Strategy: Step 1: suppose that f−1(s2,1)=f−1(s2,2), and see that S2,1=S2,2; Step 2: suppose that S2,1=S2,2, see that f−1(s2,1)=f−1(s2,2).

Step 1:

Let us suppose that f−1(S2,1)=f−1(S2,2).

f∘f−1(S2,1)=f∘f−1(S2,2). But f∘f−1(S2,1)=S2,1 and f∘f−1(S2,2)=S2,2, by 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. So, S2,1=S2,2.

Step 2:

Let us suppose that S2,1=S2,2.

f−1(S2,1)=f−1(S2,2) is immediate.

References
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