Question 7704: suppose G is the set of all functions from ZtoZ with multiplication defined by composition, i.e,f.g=fog.show that f has a right inverse in G IFF F IS SURJECTIVE,and has a left inverse in G iff f is injective.also show that the setof al bijections from ZtoZis a group under composition.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Suppose G is the set of all functions from Z to Z with multiplication defined by composition, i.e,f.g=fog. show that f has a right inverse in G IFF F IS SURJECTIVE,and has a left inverse in G iff f is injective. also show that the setof al bijections from ZtoZis a group under composition.
Let id be the identity map of Z.
1)show that f has a right inverse in G IFF F IS SURJECTIVE
proof: (==>) if g is a right inverse of f, then fog = id.
For any x in Z, we have fog(x) = x. Hence,
x = f(g(x)),this means x is the image of g(x) under f.
And so f is surjective.
(<==) If f is surjective, then for any y of Z, y =f(x) for
some x in Z. Define g:Z-->Z by g(y) = x
(choosing a unique one such x with f(x) = y).
Then, fog(y)= f(x) = y for all y in Z.
This shows fog = id.
2)f has a left inverse in G iff f is injective
proof: (==>) if g is a left inverse of f, then gof = id.
If f(x)= f(y) for x,y in Z, then
gof(x)= gof(y) --> id(x) = id(y) --> x = y
Hence, f is one-to-one (injective)
(<==) If f is injective,
y = f(x).Define g: f(Z) --> Z by g(y) = x if y = f(x).
Since if y1= f(x1), y2 = f(x2) and y1=y2
--> f(x1) = f(x2) -->x1 =x2 .So, g is a well-defined function.
Clearly ,we have gof(x) = g(f(x)) = x by definiton of g.
3) Check the 4 requirements about the defintion of groups. You will see.
The proof is straightford.
Kenny
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