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1. What is the Difference Between Range and Co domain of Function ?

2.If i say a function is one to one , onto , bijection what does it actually tell about the function is there any significance or they are just types of function ?

3. when i say $fog(x)$ then $fog(x) = f(g(x))$ we know that $g(x)$ need not be onto i understand why but why its must that f should be onto. One reason could be: if f is onto then the range of $fog(x) =$ $\text{function f}$ ,   but that doesn't make sense because say if in co domain of f there is an element which doesn't have any pre image in domain then what's the problem because we can never attain that image because there exist no pre image so how does it effect its range ?

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1. What is the Difference Between Range and Co domain of Function ?

Range is the actual values that are images of Domain. Whereas Co-Domain is  everything that you have on right side of any function.

For example, when we write $f : A \rightarrow B$ then A is Domain, B is Co-domain and the Set $R = \left \{ f(x) \,|\,x \in A \right \}$ is Range.

Say, $f : \mathbb{R} \rightarrow \mathbb{R}$ and $f(x) = e^x$ then here $\mathbb{R}$ is Co-domain and $(0, ∞ )$ is Range of $f$.

2.If i say a function is one to one , onto , bijection what does it actually tell about the function is there any significance or they are just types of function ?

Of course there is significance. In Mathematics, Functions are one the Most important concepts. To see just a glimpse of their (injection, bijection, surjection etc) significance.. refer here : https://gateoverflow.in/216802/set-theory

when i say fog(x) then fog(x) = f(g(x)) we know that g(x) need not be onto i understand why but why its must that f should be onto.

Who said that for $fog$ to be defined, $f$ must be Onto function?? It's not necessary. Neither $f$ nor $g$ have to be onto for $fog$ or $gof$ to be defined.

by Boss (26.1k points)
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Ohk sir if i say in fog(X): fog is onto then f must be onto else fog wont be onto because some elements would then always exist in co domain of fog for which there is no pre image.
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Sir so you are saying Range is subset of codomain that means that unless the function is onto we cant attain all the values of codomain right sir ?
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Sir then what is the necessary and sufficient condition for fog and gof to exist. except that the range of inner function should be the subset of domain of outer function. Is there any more condition to exist for fog and gof to exist.
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Sir According to me "the range of inner function should be the subset of domain of outer function" is the only condition because :-

Well i have drawn the cases in my copy where i took f to be not one to one and g to be not onto nad vice versa but still fog existed.
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If the function gof: A → C is surjection and g is an injection then the function f is a surjection. Sir its also correct right ?
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I think it is wrong .

F need not be surjection.

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But sir here fog which you've drawn is not onto because it contains an element in the codomain which doesn't have any pre image. But in question its said if gof is onto .....
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Ohk sir if i say in fog(X): fog is onto then f must be onto else fog wont be onto because some elements would then always exist in co domain of fog for which there is no pre image.

Absolutely right.

Sir so you are saying Range is subset of codomain that means that unless the function is onto we cant attain all the values of codomain right sir ?

Yes. Right.

Sir then what is the necessary and sufficient condition for fog and gof to exist. except that the range of inner function should be the subset of domain of outer function. Is there any more condition to exist for fog and gof to exist.

No. These conditions will suffice.

If the function gof: A → C is surjection and g is an injection then the function f is a surjection. Sir its also correct right ?

Check your first comment. If $gof$ is Surjection then g has to be Surjection.

imagine an circle,co domain is the complete circle and range is a part of it.
by (115 points)