why is the domain of $g_2 \circ g_1$ is $g^{-1}(V)$ instead of just $U$ ? (Munkres, p199)

Question

In the book of Analysis on Manifols by Munkres, at page 199, it is given that

I do not understand why the author uses $g^{-1} (V)$ as the domain of $g_2 \circ g_1$. I mean, aren't we only know that $g(U) \subseteq V$. If so, why is the author not just define $g_2 \circ g_1$ on $U$.

I mean it looks like a minor detail, but it is as if I'm missing something.

Answer 1

It may be that $g_1(U)\not\subseteq V$. In that case, there are $p\in U$ such that $g_1(p)\notin V$, which means that $p$ can't be in the domain of $g_2\circ g_1$. Thus in order to guarantee that $g_2\circ g_1$ is well-defined, we must restrict the domain to $g_1^{-1}(V)$.

Answer 2

You have two functions $f:A\rightarrow B$ and $g:C\rightarrow D$.

You want to compose these two functions and write $g\circ f :A\rightarrow D$ but you are not given that $B\subseteq C$ in which case you can write the composition...

You are not even given $f(A)\subseteq C$ in which case you can write the composition..

So, what you do is consider $f(A)\cap C$. You can call it $B'$ if you want. We have $B'\subseteq B$ and $B'\subseteq C$.

Only problem here is that, it is $f$ does not map $A$ to $f(A)\cap C$. So, we need to restrict our domain to say $C'$ such that $f(C')\subseteq f(A)\cap C$.

As $C'\subseteq A$, we have $f(C')\subseteq f(A)$. So, this only asks you for the condition $f(C')\subseteq C$ i.e., $C'\subseteq f^{-1}(C)$.

This is what they did it there...

Your case is $g_1:U\rightarrow \mathbb{R}^n$ and $g_2:V\rightarrow \mathbb{R}^n$.. You want to write $g_2\circ g_1$. So, for above reason, you should retrict domain of $g_1$..

They considered $C'=f^{-1}(C)$ and in your case it is $g_1^{-1}(V)$...