Gof is injection, does this imply that f is an injection

  • Thread starter Dustinsfl
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In summary: Since you are given that g \circ f : A\to C is injective, you need to show that f(a_1) = f(a_2) implies a_1=a_2. This is what you have to prove: that f(a_1) = f(a_2) \implies a_1=a_2. Some math books define injective as: A map f:A\to B is injective if f(a_1) = f(a_2) implies a_1 = a_2. This is what you have to prove: that f(a_1) = f(a_2) \implies
  • #1
Dustinsfl
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Let A,B, and C be nonempty sets and lef f map A to B and g map B to C.

If g of f maps A to C is an injection, then f is an injection.

Not sure how to do this proof
 
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  • #2


It seems like you always start out your posts with something like "Not sure how to do this proof" without attempting it. Why don't you assume f is NOT an injection and see if it leads to a contradiction with g of f being an injection?
 
  • #3


Of course, because if I knew how to start it or do it, I wouldn't ask.
 
  • #4


Dustinsfl said:
Of course, because if I knew how to start it or do it, I wouldn't ask.

Sure, if you knew how to do it you wouldn't need to ask. But could you thrash around with attempts a little? Other people do.
 
  • #5


You are making an assumption that I haven't. I don't post a futile attempt at it that doesn't go anywhere. With computations for classes like DE and Calc, posting the work allows others to identify errors. I can post work associated with a proof that leads no where so whomever helps would have to start from scratch anyways.
 
  • #6


Dustinsfl said:
You are making an assumption that I haven't. I don't post a futile attempt at it that doesn't go anywhere. With computations for classes like DE and Calc, posting the work allows others to identify errors. I can post work associated with a proof that leads no where so whomever helps would have to start from scratch anyways.

I'm not making the assumption you haven't done anything. I'm just saying you don't tell us what you've tried. That is not valueless even if it didn't work. Check out post 2. Try a proof by contradiction.
 
  • #7


Dustinsfl said:
Of course, because if I knew how to start it or do it, I wouldn't ask.

More often then not, my students do not know how to start because they are forgetting definitions. One tip I always give my students is to simply write down the definitions of the key words in the problem if you do not know how to start. The is almost always the first thing to try regardless of the problem, and regardless of whether you know how to start.

Since you are asked to show [tex]f[/tex] is injective, you need to know what "injective" means. Write down exactly what it means for a function to be injective.

Some math books define injective as: A map [tex]f:A\to B[/tex] is injective if [tex]f(a_1) = f(a_2)[/tex] implies [tex]a_1 = a_2[/tex]. This is what you have to prove: that [tex]f(a_1) = f(a_2) \implies a_1=a_2[/tex]

What are you given? You are given that [tex]g \circ f : A\to C[/tex] is injective. What does this mean according to the definition of "injective"? (see the definition above, but apply it to [tex]g\circ f[/tex] as opposed to just [tex]f[/tex]). It may help to switch from [tex]g\circ f[/tex] notation to [tex]g(f(x))[/tex] notation.

Can you then use the given information to deduce what you need to show?
 

1. Is "Gof" the same as "f"?

No, "Gof" and "f" are not the same. "Gof" is a composition of two functions, g and f, while "f" is just a single function.

2. What does it mean for f to be an injection?

An injection, also known as a one-to-one function, is a function where each element in the domain maps to a unique element in the codomain. This means that no two elements in the domain can map to the same element in the codomain.

3. Does "Gof" being an injection imply that "f" is also an injection?

Yes, if "Gof" is an injection, then "f" must also be an injection. This is because if "Gof" is an injection, then every element in the domain of "f" must map to a unique element in the codomain of "f".

4. Can "Gof" be an injection if "f" is not an injection?

No, "Gof" cannot be an injection if "f" is not an injection. This is because if "f" is not an injection, then there must be at least two elements in the domain of "f" that map to the same element in the codomain of "f", which would violate the definition of an injection.

5. How can I prove that "f" is an injection?

To prove that "f" is an injection, you can use the definition of an injection and show that for every pair of elements in the domain of "f", they map to different elements in the codomain of "f". You can also use techniques such as the horizontal line test or the method of contradiction to prove that "f" is an injection.

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