edited by
8,782 views
37 votes
37 votes

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
edited by

8 Answers

Best answer
36 votes
36 votes
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).
edited by
33 votes
33 votes

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  ​​​​​​

15 votes
15 votes

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.

7 votes
7 votes

Solve this question by simplifying statements as above.

Answer:

Related questions

31 votes
31 votes
7 answers
1
44 votes
44 votes
10 answers
2