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| Composition of two functions or Product of two functions |
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| Let f:AgB and g:AgB be two functions. Thus the composition of two functions f and g denoted by gof or fog is the function from A into C defined by gof = {(a,b) for some c
Î B, (a,c) Î f and (c,b)
Î g}. |
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| Remark: |
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| gof is defined only if Rf = Dg. |
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| Example 1: |
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| Suggested answer: |
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| Example 2: |
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| If f : R→R
and g: R→R denoted by f(x) = x + 2, and g(x) = 2x2 + 7. Compute
fog and gof. |
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| Statement: |
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f : S  T, g : T  U and h : U  V be function then |
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| a) (hog)of = ho(gof). |
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| b) If each f and g is one-to-one then so is gof. |
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| c) If each f and g is onto then so is gof. |
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| Proof: |
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| a) (hog)of = ho(gof) |
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Since, (hog)of and ho(gof) are functions from S to V. We have to prove that for any  the image of S under ho(gof) = image of x under (hog)of. |
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| b) If each f and g is one-to-one, then so is gof. |
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| Hence gof is one-one . |
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| c) If each f and g is also then so is gof. |
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To prove gof is onto we have to prove that every element of u
Î U is an image element for some x
Î S under gof, since g is onto 
t Î T such that g(t) = u. |
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| gof(x) = g(f(x)) = g(t) = u |
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| gof : S g U is a function . |
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| gof : S g U is one-one and onto by (b) and (c). |
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(gof)-1 exists and (gof)-1 : U®S is also a
one-one and onto function. |
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| Again f-1and y-1 exists and one-one and onto because f and g are one-one and onto by (b) and (c). |
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 (  if is one-one and onto) |
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| Further, we have |
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= f-1 (t) = x ...(ii) |
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| From (i) and (ii), we get |
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| Corollary: |
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| The function f : A g B is one-one and onto. |
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| Proof: |
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| f : A g B and is one-one and onto. |
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| \ for x Î A , we have an element b Î B such that |
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| This proves that f is one-one and onto. |
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