Exercise 7: Consider the function $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = 2x$. As a set of pairs, this can be described by
$$\{\ldots,(-2,-4), (-1,-2), (0,0), (1,2), (2,4),\ldots\}.$$
If we simply switch the inputs with outputs:
a) Is the result a function?
b) Is the result a function from $\mathbb{Z}$ to $\mathbb{Z}$?
My solution is part a):
No this is not the solution of the function it is the solution of function $Z=x/2$
My solution of part b):
(No this is the relation of $\mathbb{Z}$ to $\mathbb{Z}$)
(Can you pls check my solutions of above part)?
And now I'm confused on second part
Let $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = 2x$.
a) Find function $g: \mathbb{Z} \to \mathbb{Z}$ such that $g\circ f=\mathrm{id}_\mathbb{Z}$.
b) Show that the function from a) does not satisfy $f\circ g=\mathrm{id}_\mathbb{Z}$.
A corrected awnser to the first part:
a: This result is indeed a function $g: \{2z: z\in\mathbb{Z}\}\longrightarrow\mathbb{Z}$ because it only ever takes in even numbers. So the awnser to b is No, because the domain is not the same as $\mathbb{Z}$
Tip: you can use LaTeX notation in asking a question.
EDIT: Sorry I read the question incorrectly. Here is the awnser to the second part:
a: The anwser is the function $$g:\{2z:z\in\mathbb{Z}\}\longrightarrow\mathbb{Z}$$ $$2z\longmapsto z$$ because for any $z\in\mathbb{Z}$ it follows that: $$(g\circ f)(z)=g(f(z))=g(2z)=z$$
b: The function $f\circ g$ has a domain of $\{2z:z\in\mathbb{Z}\}$ so it can never be the identity on $\mathbb{Z}$.