What functions $f: A \to B$ and $g: B \to A$, satisfy a restriction such that $f$ is not invertible but $f \circ g=id_B$?

Question

I am caught up on the notation of $id_B$. I'm thinking that $f=x^2$, or something along those lines, but not so sure as to what $g$ may be.

Answer 1

Per our chat discussion:

Let $A = \mathbb R$, $B = \mathbb R_{\ge 0}$, and define $f: A \to B$ by $x \mapsto x^2$, and $g: B \to A$ by $x \mapsto \sqrt x$. Then $f \circ g$ is the identity function on $B$, but $f$ is not one to one.

Answer 2

Hint: If $f\circ g=\operatorname{id}_B$ then $g$ in injective and $f$ is surjective.

Answer 3

Hint 1: Since $f \circ g$ is onto, $f$ must be onto.

Hint 2: If $f: A \to B$ an onto function, try to prove that there exists some $g: B \to A$ such that

$$f(g(x))=x \,.$$

Note that for each $x$, you have $g(x)= ??$ means $f(??)=x$...

Answer 4

Assume that $B\subset A$, $B\ne A$, choose $b$ in $B$ and define $g(x)=f(x)=x$ for every $x\in B$ and $f(x)=b$ for every $x$ in $A\setminus B$. Then $f\circ g=\mathrm{id}_B$ but $f$ is not invertible.

Answer 5

Let$$f(x)=|x|,g(x)=x$$ where $A=\mathbb{R},B=\mathbb{R}_+$.