In the book of Analysis on Manifolds by Munkres, at page 202, it is asked that
Let $A$ be open in $\mathbb{R}^k$; şet $f:A\to \mathbb{R}$ be of class $C^r$. Show that the graph of $f$ is a $k-$manifold in $\mathbb{R}^{k+1}$.
What I have done is that
Define $\alpha:\mathbb{R}^k \to \mathbb{R}^{k+1}$ by $\alpha(x) = (x,f(x))$, then it is clear that $\alpha\in C^r$ is bijective and $D\alpha$ is nonsingular for $x\in A$. However, to show the continuity of $\alpha^{-1} : \mathbb{R}^{k+1} \to \mathbb{R}^k$ by $$\alpha(x_1,..,x_k,f(x_1,...,x_k)) = (x_1,...,x_k)$$, I took a open set $C$ in $R^k$, but then realised that the domain of $\alpha^{-1}$ is of measure zero in $\mathbb{R}^{k+1}$, and it is non-empty, so it is neither open nor contained any open set in $\mathbb{R}^{k+1}$, so $\alpha (C)$ cannot be open, hence $\alpha^{-1}$ cannot be continuos, which feels like as if I'm missing something that is right in front of my eyes.
Therefore, my question is that is my reasoning correct ? If not, what am I missing in here ?
Your mistake is claiming that $\alpha$ is bijective. In fact, $\alpha$ is pretty far from bijective, as its domain is only the graph of the function $f$. Therefore, for any $x\in\mathbb R^k$, the element $(x, f(x))$ is in the domain of $\alpha$, but all other elements of the form $(x, c)$ where $c\in\mathbb R$ and $c\neq f(x)$ are not in the domain.