Answer to Question #267822 in Discrete Mathematics for demkm

Question

2.     Indicate if the following statements are possible. Justify your answer. If the answer is “yes”, give a specific example of the functions. Let f:   and g: Z be two functions. Use a diagram if that helps explain the function.

a)  Is it possible that f is not onto and g ◦ f is onto? 

b)  Is it possible that g is not onto and g ◦ f is onto?

c)  Is it possible that g is not one-to-one and g ◦ f is one-to-one? 

Answer 1

Since in question domain and codomain of both functions were not exactly specified, answer will be given in assumption of domain and codomain of both functions being subsets of $Z$.

a)Yes. Since codomain of g ◦ f function is same as codomain of function g due to definition of function composition. If $f(x): A \mapsto B$ and $g(y) : B\mapsto C$ then $g ◦ f(x) : A\mapsto C \ and \ g ◦ f(x)=g(f(x))$.

For example if f:$Z \mapsto Z$ f(x)=|x| and g:$Z \mapsto Z^+$g(x)=|x|. This will result in function $g ◦ f(x) : Z\mapsto Z^+$ and obviously for every element y in Z⁺ exist at least one element x in Z such as g ◦ f(x)=y

b)No. As said above function g defines codomain of composition, and if where is not empty subset $C' \subset C$ such as $\nexists y\in B \mid g(y) \in C'$.

c)Yes. if for example f:$Z^+ \mapsto Z^+$ f(x)=2x and g:$Z \mapsto Z^+$ g(x)=|x|. g is not on-to-one function, since exists x₁=1 and x₂=-1 which make statement $g(x_1)=g(x_2) \to x_1=x_2$ false. But function g ◦ f is $Z^+ \mapsto Z^+$function and now for every x₁ and x₂ $g ◦ f(x_1)=g ◦ f(x_2) \to x_1=x_2$