Uniqueness in Munkres' proof of the Implicit Function Theorem (Analysis on Manifolds) (I'll put the whole proof here as image. I know it could be bad for future searching but I think I referenced well enough in the title)




My problem is just in the uniqueness part. 
First: it says since $g_0$ is continuous, there is neighborhood $B_0$ of $a_0$ contained in $B$ such that $g_0$ also maps $B_0$ into $V$. I need to know wheter my reasoning for understanding this is right: choose $B_0 = g_0^{-1}(V_{g_0(a_0)})\cap B$, where $V_{g_0(a_0)}$ is a neighborhood of $g_0(a_0)$ contained in $V$ and since $g_0$ is continuous, $g_0^{-1}(V_{g_0(a_0)})$ is open and thus $B_0$ as defined is open and is mapped by $g_0$ into $V$. Is that right? There is a simpler or more direct way to see this (I might be asking this because of the comment "this is easy")?
Second: I don't see where they use this fact. The fact that they agree on $B_0$ wouldn't be enough for the conclusion?
Thanks in advance
 A: If a function is continuous at a point, then it is continuous on an open neighborhood. As $g_0$ is continuous at $a_0$ it is therefore continuous on an open neighborhood of $a_0$. Call this open set (or "neighborhood") $B_0$.
Then decompose $$ B = \{ x : |g(x) - g_0(x)| = 0 \} \cup \{ x : |g(x) - g_0(x)| > 0\}. $$
By continuity of $g, g_0$ we have both those sets are open. Clearly these sets are disjoint. Hence they form a "separation" of B. By hypothesis, B is connected; therefore one of  the sets is empty. By construction we have $$ a_0 \in \{ x : |g(x) - g_0(x)| = 0 \}, $$ hence this set is non-empty. Therefore $$ \{ x : |g(x) - g_0(x)| > 0\} = \emptyset $$ which gives that $$ B = \{ x : |g(x) - g_0(x)| = 0 \}. $$ By positivity of metrics, for any $x \in B$ we get that $|g(x) - g_0(x)| = 0 \Rightarrow g(x) = g_0(x). $ Since $g = g_0$ we may conclude that the function g is unique.
I just re-formulated munkres in my own words; I hope you find this helpful.
A: Regarding the first point, you are right - this is the basic argument. I think Munkres says it is easy because this is almost the definition of continuity.
The second point is more subtle. It is not enough to say "now that $g$ and $g_0$ agree on $B_0$, we can replace $B$ with $B_0$ and be done" for the following reason: Suppose we now have a third function $g_1$. The argument above now only shows that $g_0$ and $g_1$ agree on some even smaller set $B_1$. And by considering a sequence of functions $g_i$, it may end up that they only agree on the intersection of a decreasing sequence of sets $B_i$ which may only be the point $a$. Thus we need to show that $g$ and $g_0$ agree on an open set that does not depend on the choice of $g_0$. So I think the argument that Munkres makes is necessary.  
