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# Question about proving surjective/injective functions Watch

1. Problem solved, no more help needed
2. There aren't any functions of functions here. means the function such that - i.e. function composition.
3. I assume you mean g: B -> C

Assume f is not injective, so there are distinct a, b in A such that f(a)=f(b). What is g o f (a)? How about g o f (b)? What does this mean for g o f?
4. (Original post by Hopple)
I assume you mean g: B -> C

Assume f is not injective, so there are distinct a, b in A such that f(a)=f(b). What is g o f (a)? How about g o f (b)? What does this mean for g o f?
ah yes thanks , (i) completed with that hint
i'm still not sure about (ii) however, i can understand that if g o f is surjective then f must have a range of (-infinity, infinity) which covers the requirement where g must be able to take any input b, where b(is a member of)R , am i missing a key formula/identity which explains this?
5. Firstly, the range need not be (-infinity, infinity). A could be {1,2,3,7}, B could be {2,3,4,8} and C could be {4,5,6,10} with f(x)=x+1, and g(x)=x+2. There needn't even be an algebraic map, nor the sets be of numbers.

This proof is more by definition than the last, what does g o f surjective mean about any element in C, and elements in A? What does f do to those elements in A?
6. (Original post by Hopple)
Firstly, the range need not be (-infinity, infinity). A could be {1,2,3,7}, B could be {2,3,4,8} and C could be {4,5,6,10} with f(x)=x+1, and g(x)=x+2. There needn't even be an algebraic map, nor the sets be of numbers.

This proof is more by definition than the last, what does g o f surjective mean about any element in C, and elements in A? What does f do to those elements in A?
hmm g o f surjective would mean that every element in C is related to at least one element in A,
but then the wording gets messy...
and since f:A->B , g:B->C must also be surjective (i can understand it in my head, but how to word it..!)
7. Well, g o f surjective means for any c in C, there's an a in A that is mapped to it. But then f means there's a b in B that is f(a). Then g(b) = c, and you've proven surjectivity of g.

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