If g and f o g are onto(Surjective), is f onto(Surjective)?

  • Thread starter LittleTexan
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In summary, the conversation is about a person needing help with a proof in their Discreet math course. They are trying to prove that if g and fo g are onto, then f must also be onto. They understand that for a function to be onto, it must map to all images, and they believe that since C is the image of f, f must be onto. They are seeking advice on how to prove this and for general tips on proofs. It is stated that the image of fg is a subset of the image of f, and if f is not surjective, then fg cannot be surjective.
  • #1
LittleTexan
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Hello Members

I am having a little bit of a problem solving this proof for my Discreet math course.

If g and f o g are onto(Surjective), is f onto(Surjective)? Need to prove. I believe that f has to be Onto.

so I have g: A -> B
f: B -> C

Well I understand that a function is onto(Surjective) when it maps to all images. So for g all the elements in A map(hit) element in B.

So f o g: A -> C where every element of C must be map to. I believe that since C is the Image of f that this means that f must be onto(Surjective).

Can someone give me advise on how to prove this? Or even just some advise in general on proofs.

TIA
 
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  • #2
The image of fg is a subset of the image of f. If f is not surjective then fg cannot possibly be surjective.
 
  • #3



Yes, you are correct in your understanding. Since g and f o g are onto, this means that every element in C is being mapped to by f o g, which is a composition of two functions. This also means that every element in B is being mapped to by g, and therefore every element in C must also be mapped to by f. This is because f is the second function in the composition and it is mapping the elements from B to C.

To prove this, you can use a proof by contradiction. Assume that f is not onto. This means that there exists an element in C that is not being mapped to by f. However, since g and f o g are onto, this element must be mapped to by g. But this contradicts the fact that f is not onto, because g is the first function in the composition and it is mapping all elements from A to B. Therefore, our assumption is false and f must be onto.

In general, when proving statements in mathematics, it is important to carefully understand the definitions and properties involved. In this case, understanding the definitions of onto and composition of functions is crucial. Additionally, using proof techniques such as proof by contradiction can help to logically and rigorously prove the statement. Keep practicing and seeking help when needed, and you will become more confident and proficient in proofs.
 

1. What is the definition of onto (Surjective)?

Onto (Surjective) is a term used in mathematics to describe a function that maps all elements of its domain to its codomain. In other words, every element in the codomain has at least one preimage in the domain.

2. Why is it important for a function to be onto (Surjective)?

An onto (Surjective) function ensures that every element in the codomain is being used, making it easier to solve equations or problems involving that function. It also allows for the inverse function to exist, which is useful in many mathematical applications.

3. How do you determine if a function is onto (Surjective)?

A function is onto (Surjective) if every element in the codomain has at least one preimage in the domain. This can be checked by looking at the graph of the function or by using the vertical line test. If every vertical line intersects the graph at least once, the function is onto (Surjective).

4. If g and f o g are onto (Surjective), is f onto (Surjective)?

Yes, if g and f o g are onto (Surjective), then f must also be onto (Surjective). This is because the composition of two onto (Surjective) functions will also be onto (Surjective).

5. Can a function be onto (Surjective) if its domain and codomain are not the same size?

Yes, a function can still be onto (Surjective) even if its domain and codomain are not the same size. The important factor is that every element in the codomain has at least one preimage in the domain, not the size of the sets.

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