GATE IT 2005 | Question: 31

Question

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 $g$?

• A. $g$ is onto $\implies$ $h$ is onto
• B. $h$ is onto $\implies$ $f$ is onto
• C. $h$ is onto $\implies$ $g$ is onto
• D. $h$ is onto $\implies$ $f$ and $g$ are onto

Comments

Let's see if C is true or not. Once you get C correct you will be able to see why all others are wrong.

So c is h is onto -> g is onto

Now h is onto and h is a function from A to C and all elements of C must have a pre image in A.

h is onto this is true then g is onto this needs to be false to prove the option c as wrong.

So can g be not an onto function?

Let's see,

G means onto means every element in set C must have a pre image in B. To prove this wrong let's introduce an element X in C such that it has no pre image. But this can't happen as if X has no pre image for function G. Then it means in H we have a value X that doesnot have any pre image. But this isn't possible as h is onto. So yes G can't be not onto

Hence we can't prove X -> Y as false. Therefore the statement is true.

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Similar question: https://gateoverflow.in/1168/gate-cse-2005-question-43

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All possible Variations:

h = fog

1. h onto ⇒ f is onto

2. h one to one ⇒ g is one to one

3. h is bijective ⇒ g is one to one and f is onto

4. h one to one does not imply f is one to one

5. h onto does not imply g is onto

6. Given f is bijective:
   g is one to one ⟺ fog is one to one
   g is onto ⟺ fog is onto
   g is bijective ⟺ fog is bijective

7. Given g is bijective:
   f is one to one ⟺ fog is one to one
   f is onto ⟺ fog is onto
   f is bijective ⟺ fog is bijective

8. fog is one to one and g is onto ⇒ f is one to one

9. fog is onto and f is one to one ⇒ g is onto

10. fog is bijective and g is onto ⇒ f, g both are bijective

11. fog is bijective and f is one to one ⇒ f, g both are bijective

Answer 1

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).

Comments

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

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@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.

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@Arjun Sir, in you example h(x) should be x-3? Please confirm?

Answer 2

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  ​​​​​​

Comments

Any Properties For "fog(x)". Like This...? So Please Provide.

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For the second point, I want to add that g will have to be injective on the range of f (but since it can collapse elements outside of that, hence g need not necessarily be one-one)

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@Vasu Monpara These are all the possible variations

h = fog

1) Given f is bijective:

g is one to one  ⟺ fog is one to one

g is onto ⟺ fog is onto

g is bijective ⟺ fog is bijective

2) Given g is bijective:

f is one to one ⟺ fog is one to one

f is onto ⟺ fog is onto

f is bijective ⟺ fog is bijective

3)  fog is one to one and g is onto ⇒  f is one to one

4)  fog is onto and f is one to one ⇒ g is onto

5)  fog is bijective and g is onto ⇒ f, g both are bijective

6)  fog is bijective and f is one to one ⇒ f, g both are bijective

Answer 3

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.

Comments

if h is onto then g must be onto but the converse need not be true.

Answer 4

Solve this question by simplifying statements as above.

Comments

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

Answer 5

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}

Comments

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

Answer 6

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