# Proof about injectivity

## Summary:

I need help checking my proofs as I self-study math.

I typed this up in Overleaf using MathJax. I'm self-studying so I just want to make sure I'm understanding each concept. For clarification, the notation f^{-1}(x) is referring to the inverse image of the function. I think everything else is pretty straight-forward from how I've written it. Thank you for your help!

## Answers and Replies

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Math_QED
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You must show that ##x_1 \in f^{-1}(f(E)) \iff x_1 \in E##. No need to introduce ##x_2##. I see why you did that: to get to the definition of injectivity, but it is confusing.

Notice first that one inclusion always holds, without injectivity. Indeed, ##f^{-1}(f(E))## is the set of all elements in ##A## that get mapped to ##f(E)##. Clearly ##E## is contained in this set, so we always have (even without injectivity) ##E \subseteq f^{-1}(f(E))##.

For the converse, the injectivity will be crucial.

If ##x \in f^{-1}(f(E))##, then you know that ##f(x) \in f(E)##. How can you proceed and use the injectivity?

PeroK
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Your proof doesn't make much sense to me. When are two sets equal?

You must show that ##x_1 \in f^{-1}(f(E)) \iff x_1 \in E##. No need to introduce ##x_2##. I see why you did that: to get to the definition of injectivity, but it is confusing.

Notice first that one inclusion always holds, without injectivity. Indeed, ##f^{-1}(f(E))## is the set of all elements in ##A## that get mapped to ##f(E)##. Clearly ##E## is contained in this set, so we always have (even without injectivity) ##E \subseteq f^{-1}(f(E))##.

For the converse, the injectivity will be crucial.

If ##x \in f^{-1}(f(E))##, then you know that ##f(x) \in f(E)##. How can you proceed and use the injectivity?
Thank you for the reply! I always forget that to show equality after I have to clearly show that each set is a subset of the other :/
The only next step I can think of is that if f(x) is in f(E), then x is in E, right?

Math_QED
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The only next step I can think of is that if f(x) is in f(E), then x is in E, right?
Why? Is this true without injectivity?

Why? Is this true without injectivity?
Well if the function is injective then we know that x maps to at most one y = f(x). So I don't think this could be true without inectivity.

Math_QED
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Well if the function is injective then we know that x maps to at most one y = f(x). So I don't think this could be true without inectivity.
You get the idea, but at this point it is important to write down everything in huge detail!

What is the definition of ##f(E)##? Thus what does ##x \in f(E)## mean?

The definition of f(E) is {f(x) : x is in E}.

Math_QED
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The definition of f(E) is {f(x) : x is in E}.
Okay, now answer the second question. What does ##x\in f(E)## mean?

For x to be an element of f(E) it would have to be equal to some f(x) in f(E), right? Just based on the definition. I'm sorry if I'm being slow.

Math_QED
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For x to be an element of f(E) it would have to be equal to some f(x) in f(E), right? Just based on the definition. I'm sorry if I'm being slow.
Yes, so we left of at ##f(x)\in f(E)##, so there is an ##e\in E## (do not call this new variable ##x##, this name is already used!) with ##f(x)=f(e)##.

Can you conclude?

Yes, so we left of at ##f(x)\in f(E)##, so there is an ##e\in E## (do not call this new variable ##x##, this name is already used!) with ##f(x)=f(e)##.

Can you conclude?
Well if f(x) = f(e) then x = e.

Math_QED
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Well if f(x) = f(e) then x = e.
Yes, so is ##x \in E## true?

Yes!

Math_QED
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Yes!
Congrats, now your proof works!

Congrats, now your proof works!
Wow, I really overcomplicated things XD

Math_QED
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Wow, I really overcomplicated things XD
Well, let's check if you learned from it! Show the following statement (if you want to/have the time):

If ##f## is surjective, then ##f(f^{-1}(E))=E##.

PeroK
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Congrats, now your proof works!
Are you sure?

Math_QED
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Are you sure?
I meant that the OP has now a working proof (not the one he wrote down originally):

If ##x \in f^{-1}(f(E))##, then ##f(x) \in f(E)##, so there is ##e\in E## with ##f(x) = f(e)##. By injectivity, ##x=e\in E##. This shows ##f^{-1}(f(E)) \subseteq E##, the other inclusion is trivial.

PeroK
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Wow, I really overcomplicated things XD
An alternative method is to show

1) first that for any set ##E## and any function ##f## we have ##E \subseteq f^{-1}(f(E))##

2) If ##E \subset f^{-1}(f(E))##, then ##f## is not one-to-one.

Math_QED
Math_QED
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An alternative method is to show

1) first that for any set ##E## and any function ##f## we have ##E \subseteq f^{-1}(f(E))##

2) If ##E \subset f^{-1}(f(E))##, then ##f## is not one-to-one.
I prefer the notation ##\subsetneq## instead of ##\subset##, but yes, that is a good approach too.

PeroK
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I meant that the OP has now a working proof (not the one he wrote down originally):

If ##x \in f^{-1}(f(E))##, then ##f(x) \in f(E)##, so there is ##e\in E## with ##f(x) = f(e)##. By injectivity, ##x=e\in E##. This shows ##f^{-1}(f(E)) \subseteq E##, the other inclusion is trivial.
I wasn't disputing that you had a proof!

Well, let's check if you learned from it! Show the following statement (if you want to/have the time):

If ##f## is surjective, then ##f(f^{-1}(E))=E##.
E is a subset of A or B?

Math_QED
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E is a subset of A or B?
Of ##B##, or ##f^{-1}(E)## wouldn't make sense.

Of ##B##, or ##f^{-1}(E)## wouldn't make sense.
Ah, ok. I was drawing out a picture and it wasn't making sense.