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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?

  1. $f$ and $g$ should both be onto functions
  2. $f$ should be onto but $g$ need not to be onto
  3. $g$ should be onto but $f$ need not be onto
  4. both $f$ and $g$ need not be onto
asked in Set Theory & Algebra by Boss (18.3k points)
edited by | 1.8k views
+1

This might help ...

6 Answers

+28 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.
answered by Veteran (367k points)
selected by
0
Hi @Arjun sir ,

In your given example , can you please explain the relation in the function ? I mean the reason behind

f={(0,3),(3,5),(4,4)} and g={(1,3),(2,4),(0,0)}

Can you please explain a bit.
0
It is just a random example. but there is a correction on $f$, please see the change.
+1

Hi @Arjun sir

The range of fog is always bounded  to f 's range.f ,g has got its own dom, range.In fog I m trying to put a restriction on the values f can accept which is input to f in fog.this input is also range of g.

range of fog cannot exceed beyond range of f.fog is onto means my range of fog =range of f.since both equal fog onto means f has to b onto.Is this understanding ok?

0
Yes. That is correct.
+1
Can you please explain what should be the domain for function h?
+3
Domain of h = fog is all x from domain of g such that g(x) is in domain of f.
0

@arjun sir

                                                   for f:-{   (0,5),(3,4),(4,4),(25,3)  }  

{(1,3),(2,4),(0,0)}g={(1,3),(2,4),(0,0)} (2525 in BB not mapped to by gg, hence gg is not ONTO)

then fog={(1,4),(2,4),(0,5)  }(3 in c not mapped)...

so not onto....sir check this 

+13 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.

answered by Loyal (8.7k points)
edited by
0
B has two value ... bt one of its value indicating one value of C ... how does it make onto ???
+5 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 

answered by Active (2.5k points)
0 votes
Option b.
answered by Active (3.3k points)
0 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.

answered by Loyal (7.7k points)
edited by
+1
but f : B -> C is a function.So for every x belongs to B we must have f(x)=c belongs to C
0 votes

Option (B) is correct.


answered by Active (1.1k points)
edited by
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