Respuesta :
Let f:X→Ybe a function. Then the function f is a bijection if the relation {(y ,x)∈Y×X∣ (x ,y)∈f} is a function.
Assume that X→Y is a bijective function.
i.e. f:X→Y is one-one and onto function.
so different elements in X have different images in Y
and b ∈ Y there exists a ∈ X such that f(a)=b.
R= {(y ,x)∈Y×X∣(x ,y)∈f} is a relation
we wish to prove R is a function.
R is a relation from Y to X
let Y∈Y by assumption f is bijection function from X to Y
for all y ∈ Y there exists X∈ X such that f(x)=y.
so for every element in Y there is mapping element in X.
if (x ,y) ∈ f then (y ,x) ∈ R
therefore R is a function.
let X1, X2, ∈ X then there exists some y1, y2,∈ y such that
take f ( x1 )= y1 and f( x2)=y2
by assumption R is a function from Y to X that is inverse of f is existing means f is bijection.
for bijective functions only inverse exists.
If f:X→Y is a function. Then the function f is a bijection if the relation
{(y ,x)∈Y×X∣(x, y)∈f} is a function.
To learn About bijection:
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