Given $f:A\to B$, and $g:f(A)\to A$, show that if $g \circ f = I_A$ and $f \circ g = I_{f(A)}$, then $f$ is $1-1$ and $g=f^{-1}$

Question

Given $f:A\to B$, and $g:f(A)\to A$, show that if $g \circ f = I_A$ and $f \circ g = I_{f(A)}$, then $f$ is one-to-one and $g=f^{-1}$

$I_A$ denotes the identity element in $A$ and $I_{f(A)}$ denotes the identity element of $f(A)$

I want to somehow get from $g \circ f = I_A$ to $\forall a_1, a_2 \in A, f(a_1)=f(a_2)\implies a_1=a_2$

I'm pretty lost...

Answer 1

Suppose $f(a_1) = f(a_2)$. Then $g \circ f(a_1) = g \circ f(a_2)$. But $g\circ f = I_A$ and $I_A(a) = a$ for all $a \in A$, hence $a_1 =g \circ f(a_1) = g \circ f(a_2)= a_2$. So $a_1 = a_2$. So $f$ is injective.