I found many people answered using diagrams but did not mark the blue links as shown. These blue links are necessary as f is a function,

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39 votes

Let $f: B \to C$ and $g: A \to B$ be two functions and let $h = f o g$. Given that $h$ is an onto function which one of the following is TRUE?

- $f$ and $g$ should both be onto functions
- $f$ should be onto but $g$ need not to be onto
- $g$ should be onto but $f$ need not be onto
- both $f$ and $g$ need not be onto

6

48 votes

Best answer

B. $g$ need not be onto.

Let,

$A = \left\{0, 1, 2\right\}, B = \left\{0, 3, 4, 25\right\}, C = \left\{3, 4, 5\right\}$

$f = \left\{(0,3), (3, 5), (4, 4), (25, 3) \right\}$

$g = \left\{(1,3), (2, 4), (0,0)\right\}$ ($25$ in $B$ not mapped to by $g$, hence $g$ is not ONTO)

$h = \left\{(0,3), (1, 5), (2, 4)\right\}$

Now, $h$ is an onto function but $g$ is not.

$f$ must be an onto function as otherwise we are sure to miss some elements in range of $h$ making it not onto.

Let,

$A = \left\{0, 1, 2\right\}, B = \left\{0, 3, 4, 25\right\}, C = \left\{3, 4, 5\right\}$

$f = \left\{(0,3), (3, 5), (4, 4), (25, 3) \right\}$

$g = \left\{(1,3), (2, 4), (0,0)\right\}$ ($25$ in $B$ not mapped to by $g$, hence $g$ is not ONTO)

$h = \left\{(0,3), (1, 5), (2, 4)\right\}$

Now, $h$ is an onto function but $g$ is not.

$f$ must be an onto function as otherwise we are sure to miss some elements in range of $h$ making it not onto.

0

27 votes

A function f: X → Y is called onto function if, for every value in set Y, there is a value in set X.

Given that, f: B → C and g: A → B and h = f o g.

Note that the sign o represents composition.

h is basically f(g(x)). So h is a function from set A

to set C.

It is also given that h is an onto function which means

for every value in C there is a value in A.

We map from C to A using B. So for every value in C, there must be a value in B. It means f must be onto. But g may or may not be onto as there may be some values in B which don't map to A. **Example :**

Let us consider following sets

A : {a1, a2, a3}

B : {b1, b2}

C : {c1}

And following function values

f(b1) = c1, f(b2) = c1

g(a1) = b1, g(a2) = b1, g(a3) = b1

Values of h() would be,

h(a1) = c1, h(a2) = c1, h(a3) = c1

Here h is onto, therefore f is onto, but g is

onto as b2 is not mapped to any value in A.

Given that, f: B → C and g: A → B and h = f o g.

6 votes

A fuction is called onto if each member of codomain have preimage ( means should be a edge from domain to codomain for each member of codomain )

h=fog=f(g)

so first evalute g then f

A---B---C

option A) f and g should both be onto functions : **FALSE** becoz it not neccessary that if we go from A to C we have edge for every member of B

option B) f should be onto but g need not to be onto : **TRUE**

option C) g should be onto but f need not be onto : **FALSE** same explanation as option A.

option D) both f and g need not be onto : **FALSE** becoz if f is not onto then how h will be onto .

**Hence Ans is B **

6 votes

Example to prove that g neednot be onto ...

f should be onto because f is not onto then it means there is atleast one element in C which will not have a pre-image in B. If that element in C doesnot have preimage in B, then it is impossible for it to have a preimage in A. If this happens then h will also become not onto. But in question it is given that h has to be onto.