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

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In summary, the conversation is discussing how to start proving a statement about the preimage of a singleton under a function and how it relates to the function being a bijection. The person asking for help is lost after some initial facts and is seeking assistance in proving the statement. They mention that the professor forgot the full statement, which is that if the preimage of a singleton is a singleton for all elements of the codomain, then the function is a bijection. The conversation ends with the person being prompted to prove surjectivity and injectivity to show that the function is a bijection.
  • #1
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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  • #2
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
 
  • #3
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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  • #4
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?
 
  • #5
Surjectivity and interjectivity
 
  • #6
cbarker1 said:
Surjectivity and interjectivity
Surjectivity and injectivity, yes. But what does that mean, what do you have to check?
 
  • #7
cbarker1 said:
Surjectivity and interjectivity
Let's see you prove those.
 

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

What does it mean to prove the preimage of a singleton under f is a singleton for all y?

Proving the preimage of a singleton under f is a singleton for all y means showing that for any given function f, if the input or domain is a single element, then the output or range will also be a single element.

Why is it important to prove the preimage of a singleton under f is a singleton for all y?

This is important because it helps us understand the behavior of a function and its relationship between the input and output values. It also allows us to make predictions and draw conclusions about the function.

What are some techniques used to prove the preimage of a singleton under f is a singleton for all y?

Some techniques that can be used to prove this include direct proof, contradiction, and mathematical induction. These techniques involve using logical reasoning and mathematical principles to show that the preimage of a singleton will always result in a singleton.

Can the preimage of a singleton under f be a non-singleton set?

No, the preimage of a singleton under f can only be a singleton. This is because a singleton is defined as a set with only one element, and the preimage is the set of all elements that map to a specific output. Therefore, if the output is a singleton, the preimage can only contain one element.

How does proving the preimage of a singleton under f is a singleton for all y relate to one-to-one functions?

Proving the preimage of a singleton under f is a singleton for all y is closely related to one-to-one functions. One-to-one functions have the property that each input has a unique output. Therefore, if the preimage of a singleton is a singleton, it shows that the function is one-to-one, as each input only has one corresponding output.

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