Proving the preimage of a singleton under f is a singleton for all y

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Homework Help Overview

The discussion revolves around proving a statement related to the preimage of a singleton under a function. The original poster seeks to establish that the preimage of a singleton under a function is itself a singleton, and vice versa, within the context of a function mapping from set X to set Y.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of preimages and the implications of a function being a bijection. Questions are raised about the full statement of the problem and the necessary conditions for a function to be bijective.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the original statement and seeking clarification on the definitions and requirements for bijections. Some participants have offered guidance on what needs to be shown for the proof.

Contextual Notes

There appears to be some confusion regarding the full statement of the problem, as well as the definitions of surjectivity and injectivity. The original poster expresses uncertainty about how to proceed with the proof.

cbarker1
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Homework Statement
Let #f:X \to Y# be a function. Show that if #f^{-1}({y})# is a singleton for all #y \in Y#.
Relevant Equations
Definition of Preimage is #f^{-1}(B)={ x\in X: f(x) \in B}# where B is a subset of Y.
#f^{-1}({y})={x}#
Dear Everyone,

I have some trouble how to start the proof of this statement. I need to prove the preimage of the singleton under f is the subset of singleton of x and vice versus. My attempt is this:Given y.

So we know that definition of the preimage is when all #x# is in #X# , then #f(x) \in B#.I am lost after these facts.

Thank for any assistance,

Cbarker1
 
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cbarker1 said:
Homework Statement:: Let #f:X \to Y# be a function. Show that if #f^{-1}({y})# is a singleton for all #y \in Y#.
Relevant Equations:: Definition of Preimage is #f^{-1}(B)={ x\in X: f(x) \in B}# where B is a subset of Y.
#f^{-1}({y})={x}#

Dear Everyone,

I have some trouble how to start the proof of this statement.

I'm not surprised. What is the full statement? Show if ... then?
What if I map the entire set ##X## onto a single point ##y\in Y##?

cbarker1 said:
I need to prove the preimage of the singleton under f is the subset of singleton of x and vice versus. My attempt is this:Given y.

So we know that definition of the preimage is when all #x# is in #X# , then #f(x) \in B#.I am lost after these facts.

Thank for any assistance,

Cbarker1
 
fresh_42 said:
I'm not surprised. What is the full statement? Show if ... then?
What if I map the entire set ##X## onto a single point ##y\in Y##?
The professor was typing quickly and forget then statement. If #f^{-1}({y})# is a singleton for all #y \in Y#, then #f# is bijection.
 
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cbarker1 said:
The professor was typing quickly and forget then statement. If #f^{-1}({y})# is a singleton for all #y \in Y#, then #f# is bijection.
What two things do you have to show for ##f## to be a bijection?
 
Surjectivity and interjectivity
 
cbarker1 said:
Surjectivity and interjectivity
Surjectivity and injectivity, yes. But what does that mean, what do you have to check?
 
cbarker1 said:
Surjectivity and interjectivity
Let's see you prove those.
 

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