I am reading Tu's book An Introduction to Manifolds (second edition). Specifically, Example $5.14$ in Section $5.4$ which states that
For a subset of $A\subset\mathbb{R}^n$ and a function $f:A\rightarrow\mathbb{R}^m$, the graph of $f$ is defined to be the subset $$\Gamma(f)=\{(x,f(x))\in A\times\mathbb{R}^m\}$$ If $U$ is an open set of $\mathbb{R}^n$ and $f:U\rightarrow\mathbb{R}^n$, then the two maps $$\phi:\Gamma(f)\rightarrow U,\ (x,f(x))\mapsto x,\ (\textbf{1},f):U\rightarrow\Gamma(f),\ x\mapsto (x,f(x))$$ are continuous and inverse to each other, and so are homeomorphism.
My questions are the following:
How can $f$ map points from $A$ to $\mathbb{R}^m$ and at the same time from $U$ to $\mathbb{R}^n$?
What exactly is the notation $(\textbf{1},f):U\rightarrow\Gamma(f)$? I haven't seen anything like that before...
Any help will be appreciated.
In the first sentence, Tu is defining the graph $\Gamma(f)$ of a function $f : A \to \mathbb{R}^m$ where $A$ is any subset of $\mathbb{R}^n$. In the second sentence, Tu is applying that definition to a function $f : U \to \mathbb{R}^m$ where $U$ is open and claiming that $\Gamma(f)$ is homeomorphic to $U$. (Note, there is a typo in the book. Tu writes $f : U \to \mathbb{R}^n$ when it should be $f : U \to \mathbb{R}^m$, see the discussion in the comments below.)
Here Tu is using ${\bf 1}$ to denote the identity function ${\bf 1} : U \to U$, so ${\bf 1}(x) = x$. So $({\bf 1}, f) : U \to U\times\mathbb{R}^n$ with $({\bf 1}, f)(x) = ({\bf 1}(x), f(x)) = (x, f(x))$ which actually lies in $\Gamma(f)$, the graph of $f$, so we can instead regard $({\bf 1}, f)$ as a map $U \to \Gamma(f)$.