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 $g$? $g$ is onto $\implies$ $h$ is onto $h$ is onto $\implies$ $f$ is onto $h$ is onto $\implies$ $g$ is onto $h$ is onto $\implies$ $f$ and $g$ are onto Set Theory & Algebra gateit-2005 set-theory&algebra functions normal + – Ishrat Jahan asked Nov 3, 2014 edited May 31, 2018 by Arjun Ishrat Jahan 8.8k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply smsubham commented Mar 21, 2018 reply Follow Share 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. 7 votes 7 votes Akash Papnai commented Jan 14, 2020 reply Follow Share Exact same question: https://gateoverflow.in/1168/gate2005-43 but g is changed to f and f is changed to g. 4 votes 4 votes Ashutosh777 commented Jan 24, 2021 reply Follow Share @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??? 0 votes 0 votes Overflow04 commented Oct 17, 2022 reply Follow Share @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. 0 votes 0 votes Please log in or register to add a comment.
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). Arjun answered Nov 17, 2014 edited Mar 23, 2021 by soujanyareddy13 Arjun comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Nashreen Sultana commented Jan 2, 2017 reply Follow Share 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 0 votes 0 votes Xylene commented Jul 22, 2017 reply Follow Share @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. 1 votes 1 votes Shubhgupta commented Jul 16, 2018 reply Follow Share @Arjun Sir, in you example h(x) should be x-3? Please confirm? 0 votes 0 votes Please log in or register to add a comment.
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 Warrior answered Aug 12, 2017 Warrior comment Share Follow See all 3 Comments See all 3 3 Comments reply Sona Barman commented Apr 7, 2019 reply Follow Share Very helpful information. 0 votes 0 votes chirudeepnamini commented Aug 24, 2019 reply Follow Share "If h is one-one (injective) then f is one-one".. From this statement, can we assume that g need not be one-one?? 0 votes 0 votes Vasu Patel 13 commented Aug 4, 2023 reply Follow Share Any Properties For "fog(x)". Like This...? So Please Provide. 0 votes 0 votes Please log in or register to add a comment.
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. Vicky rix answered Sep 16, 2017 Vicky rix comment Share Follow See 1 comment See all 1 1 comment reply Pranavpurkar commented Oct 19, 2022 reply Follow Share if h is onto then g must be onto but the converse need not be true. 0 votes 0 votes Please log in or register to add a comment.
7 votes 7 votes Solve this question by simplifying statements as above. jatin saini answered Jan 20, 2017 jatin saini comment Share Follow See 1 comment See all 1 1 comment reply Puja Mishra commented Jan 5, 2018 reply Follow Share Upload a gd quality picture ... its blur .... 2 votes 2 votes Please log in or register to add a comment.