While reading the proof of the inverse function theorem I stumbled on the following and I am not sure if I have comprehended the idea correctly.
Let be $g$ a function $g: V \to U$, where $U:= g(V)$ and be $f$ another function with $f: U \to V.$ Further we assume that $(f\circ g)(x)=x.$
Now I want to show that both functions are injective and that $f$ is the inverse of $g$:
1.) We take two $x,y \in V$ with $g(x)=g(y).$ It follows that $f(g(x)) = f(g(y)) \Rightarrow x=y.$ Hence, $g$ is injective.
2.) Now we take $m,n \in U$ with $f(m)=f(n)$. Then there exist two $x,y \in V$ with $g(x)=m$ and $g(y)=n.$ From $f(m)=f(n)$ it follows that $f(g(x))=f(g(y)).$ As we know that $(f\circ g)(x)=x$ we can conclude: $x=y \Rightarrow g(x)=g(y)\Rightarrow m=n.$
3.) Let be $y \in U$ so we now that there exists a $x \in V$ with $g(x) =y.$ Hence, $(g \circ f)(y) = g(f(y))=g(x)=y \Rightarrow g \circ f = y$
Are these steps proven correctly?