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

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???

---

@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.

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

---

@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?

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

Very helpful information.

---

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

From this statement, can we assume that g need not be one-one??

---

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

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 ....