0
$\begingroup$

This question is inspired by this question.

Let $f\colon S_1\rightarrow S_2$ for some topological spaces $S_1,S_2$. Suppose that for every $x_0\in S_1$ there exists an open set $U\subset S_1$, such that $\forall x\in U, f(x)=f(x_0)$.

What are the sufficient and necessary conditions on $S_1$, that $f$ is constant?

As shown in the previous question:

Necessary conditions are: $S_1$ is connected

Sufficient condition: $S_1$ is pathconnected

$\endgroup$
0
$\begingroup$

$S_1$ being connected is both, sufficient and necessary. That it's necessary has already been showed.

For the sufficient:

Let $y_0\in f(S_1)$. Consider the set $$A = \bigcup_{x\in f^{-1}(y_0)}U_x $$ Where $U_x$ is the open set containing $x$, s.t. $f|_U$ is constant. $A$ is open, as it's the union of open sets. Furthermore note $A=f^{-1}(y_0)$. A similar set can be constructed: $$B = \bigcup_{x\notin f^{-1}(y_0)}U_x$$ By construction $y_0\notin f(B)$. Furthermore $A\cup B=S_1$ and both are open. Which is a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.