This is an exercise from Tao's analysis chapter $6$.
Exercise $6.7.3$. Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be a continuously differentiable function such that $f'(x)$ is an invertible linear transformation for every $x \in \mathbb{R}^n$. Show that whenever $V$ is an open set in $\mathbb{R}^n$, that $f(V)$ is also open. (Hint: use the inverse function theorem.)
Here is the solution attached:
Proof.
Choose $\forall\; y \in f(V)$, then $\exists\; x \in V, f(x) = y$, since $V$ is open, we can find $r > 0$ s.t. $B(x, r) \subseteq V$. Also since $f'(x)$ is invertible, by the inverse function theorem we can find open sets $U, W$ s.t. $x \in$ $U, y \in W$, and $f$ is a bijection from $U$ to $W$, thus we have $f^{-1}: W \to U$ well defined. Now for $\forall\; w \in W, f^{-1}(w) \in U$ and $f'\left(f^{-1}(w)\right)$ is invertible, by the inverse function theorem again, $f^{-1}$ is differentiable at $f\left(f^{-1}(w)\right)=w$, thus continuous at $w$, so $f^{-1}$ is continuous on $W$.
Now consider $V':=B(x, r) \cap U$ which is open, let $W':=f\left(V'\right) \subseteq f(U)=W$, we see $f$ is a bijection from $V'$ to $W'$, thus $V' = f^{-1}\left(W'\right)$, so $W'$ is open. We also have $V' \subseteq V$, thus $W' \subseteq f(V)$, since $x \in U, x \in B(x, r)$, we have $x \in V'$ and $f(x) = y \in W$, this means $y$ is an interior point of $f(V)$ and thus $f(V)$ is open. $\hspace{60pt}$$\blacksquare$
The solution attached seems overly complex (but maybe it's the simplest), and I am wondering if there is any way to make it more concise and more readable.
Best Answer
The inverse function theorem states that if $O$ and $\Omega$ are open and $f : O \to \Omega$ is $C^k$, $k \geq 1$, and $Df(x_0)$ is invertible, then there are neighborhoods $A$ of $x_0$ and $B$ of $f(x_0)$ such that $f : A \to B$ is a $C^k$ diffeomorphism.
This gives an easy proof of your statement:
Let $y=f(x)∈f(V)$. By the inverse function theorem, there are neighborhoods $A⊂V$ of $x$ and $B$ of $y$ such that $f:A→B$ is a $C^1$ diffeomorphism. We have $B=f(A)⊂f(V)$. This shows that $f(V)$ is open.