Let f: A to B be function between two sets A and B Show that the following are t
ID: 3008527 • Letter: L
Question
Let f: A to B be function between two sets A and B Show that the following are true:
1- f in injective IF AND ONLY IF every b that belongs to B, the set inverse of functionof (b) contains at most one element.
2- f is surjective IF AND ONLY IF for every b belonging to B, the set inverse function of (b) is not empty
3- f is bijective IF AND ONLY IF for every b in B, the set inverse function of (b) contains exactly one element.
Question 2:
Let M be a vector space over a field K. Let X be a nonempty set. Consider the set, Hom(X,M)={f:X imples M}; i.e. the set of all functions from X into M. If f,g belong to Hom(X,M) then let f+g be the function such that (f+g)(x)=f(x)+g(x) for every x that belongs to X. If a belongs to K and f belongs to HOM(X,M), define the scale product a, f to be the function such that (a,f)(x)=a,f(X). Then Hom(x,M) is a vector space over K with the above addition and scalar product.
Explanation / Answer
1(a) Necessary condition : Let b belongs to B, then it will surely have a preimage in A because f: A-->B. Now if such function is injective or having 1-1 mapping then this preimage will be a unique preimage for the value b in its image and thus the range of inverse function of value b for set B, will have atmost one element only.
Sufficient condition : If inverse function of (b) has atmost one element only that means the domain of its inverse function has only one element that is b, and as it is a mapping so this b will also be associated with a unique element in its codomain and thus it is 1-1 mapping and thus f is injective.
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2) A function is surjective if it is ONTO or if Range of f = codomain of f or no element of codomain remains unassociated with elements of domain. Now if the set of inverse of element b in set B is not empty that means that set B is its range as well as codomain and that means that range of function f = codomain of f such f is ONTO and thus f is surjective.
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3) A function is bijective if it is both 1-1 as well as ONTO. Now as the inverse function of b has only one element that means its domain has only one element that is b and as b is associated with a unique element of its range ( because of definition of a function) and this unique element is the only element of range as well as codomain, that means it is surely 1-1 mapping here as well as as codomain has only one element that is associated with element b of domain of inveres function of b of set B, that means no element of codomain is remaining unasociated here and thus this inverse function is ONTO also and thus f is BIJECTIVE.
Proved
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