Composition of functions

Question

There exist 3 sets(A,B,C) and 2 functions f and g.

g be a function from set A to set B
f be a function from set B to set C

then composition of both the functions is denoted by $f \circ g$ which exists.

Then what is the necessary condition for the following 2 functions for the existence of $g \circ f$

a) Injection
b) Surjection
c) Bijection
d) None of these

Comments

I agree with you @Mk Utkarsh .

Let me put this more clearly, here question asks-

What is the necessary condition for the following 2 functions for the existence of g∘f?

AFAIK, we should first presume that gof exists and then find counterexamples to it.

1. Assume f and g are not injection and then make gof exist

CounterExample commented above would suffice.

2. Assume f and g are not surjection and then make gof exist

3. Assume f and g are not bijection and then make gof exist

Same as point 2.

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gof exists if inverse function of fog exists for it

Isnot it the answer for it??

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 srestha  yeah in short 

Answer 1

For simplicity let's check bijection first. If bijection can't guarantee the existence of gof then neither Injectivity nor Surjectivity can.

Let,

• Set A = {1, 2, 3}
• Set B = {4, 5, 6}
• Set C = {7, 8, 9}
• g(A->B) is as follows g(1) = 4, g(2) = 5, g(3) = 6 , which is a bijection
• f(B->C) is as follows f(4) = 7, f(5) = 8, f(6) = 9, which is also a bijection

Let's check if fog exists. So,

• f(g(1)) = f(4) = 7
• f(g(2)) = f(5) = 8
• f(g(3)) = f(6) = 9
  So, fog exists.

Let's check if gof exists. If bijection were sufficient then gof should exist.

• g(f(4)) = g(7) = not defined
• g(f(5)) = g(8) = not defined
• g(f(6)) = g{9} = not defined

So, it turns out that gof still doesn't exist. Hence, correct answer is D) None of these

Comments

yes, good :)

but then gof doesnot exists.

But here we need to prove gof exists

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@srestha : Ma'am, I get the point. We actually need to do it the other way. We need to find whether or not any of the given properties are mandatory to hold in all cases to imply existence of gof. I'll think more on it. :)

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does it mean for fog(x) and gof(x) to exist  bijection is not a necessary condition however if f and g are both bijective  functions then only fog(x) and gof(x) both will be bijection plz correct if i am wrong