Let $f$ be a function from a set $A$ to a set $B$, $g$ a function from $B$ to $C$, and $h$ a function from $A$ to $C$, such that $h(a) = g(f(a))$ for all $a ∈ A.$ Which of the following statements is always true for all such functions $f$ and $g$?
Use diagrams for understanding and solving this kind of questions
Here g is onto but h isn't,
Here h is onto so g has to be unto. For g to be not onto, any element of codomain of g shouldn't be linked to the middle set in the diagram but if that happens h also cannot be onto.
Exact same question: https://gateoverflow.in/1168/gate2005-43 but g is changed to f and f is changed to g.
@smsubham
It is given that f is a function from a set A to set B i.e. each and every element of set A is mapped to exactly one element of set B.
but in the given figure of A and B , D is not mapping to any element of B.
So given figure(except figure C) is not a function at all.
Correct me if I am wrong.
Sir, I have taken an example and it seems like option c fails in this case. Please verify and correct me if I am wrong.https://gateoverflow.in/?qa=blob&qa_blobid=14984829032615331052
f : A==>B ,g: B==>c and h :A==>C , h = gof
Properties of gof i.e h:
1. If h is Onto (surjective) then g is onto.
2.If h is one-one (injective) then f is one-one.
3.If h is bijective then g is onto and f is one-one.
Example where g is onto but h is not onto
But if h is onto, g must be onto.Because if every element in C has a preimage in A, then every element in C should also have preimage in B.
Solve this question by simplifying statements as above.
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