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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 $g$?

1. $g$ is onto $\implies$ $h$ is onto
2. $h$ is onto $\implies$ $f$ is onto
3. $h$ is onto $\implies$ $g$ is onto
4. $h$ is onto $\implies$ $f$ and $g$ are onto

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 in example B: g should be a functions from B to C right?? but in your example w is not mapped to any of the C then how is it a function???

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Let $h$ be onto (onto means co-domain = range). So, $h$ maps to every element in $C$ from $A.$ Since $h(a) = g(f(a)),\:g$ should also map to all elements in $C.$ So, $g$ is also onto.

$\implies$ option (C).
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sir , logic in this is that function f can be mapped to all the elements of set B or not so f may or may not onto

now (c) option says  h is onto => g is onto   : this  tells us that if function  h is onto than g should be onto because function g is equal to function h..?? plzz correct if i am wrong ...

thank u..

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

@Nashreen, In your example, H itself is not onto function as 14 in C is not mapped to any element in A. So option C will be F ->T which is true. So option C is correct.
@Arjun Sir, in you example h(x) should be x-3? Please confirm?

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.

## The correct answer is, (C) h is onto => g is onto ​​​​​​

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"If h is one-one (injective) then f is one-one"..

From this statement, can we assume that g need not be 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.

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Solve this question by simplifying statements as above.

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Upload a gd quality picture ... its blur ....

If condition p --> q is true then its contrapositive ( ~q --> ~p ) is also true

above example is if g is not on to then 'h' is also not onto

so its contrapositive is also true that is { h is onto --> g is onto}

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What hav u done ?? can u explain a bit ??

…....………………………………………………

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option D
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If B contains some elements, not mapped by f, (D) will be false.
it shold b f(g(a)) ...is it typo in question ?

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