4
$\begingroup$

Let $\Omega$ be a discrete topological space and $X$ any space:

which maps $f: \Omega \to X $ are continuous? which maps $f: X \to \Omega $ are continuous for each topology on $Y$ ?

Attempt:

For the first part, I believe any map $f: \Omega \to X $ is continuous since the topology of $\Omega$ is $2^{\Omega} $ and so for any open $U $ in $X$, $f^{-1}(U) \subset \Omega$ must be open.

I am stuck on the second question. Any help would be appreciated.

$\endgroup$
  • $\begingroup$ For the second part, consider these kinds of functions: the identity map, and constant functions. If $f$ is non-constant, can you devise a topology on $X$ for which $f$ is NEVER continuous? $\endgroup$ – David Wheeler Mar 3 '15 at 2:26
  • $\begingroup$ @DavidWheeler That is not true $X$ can be disconnected. $\endgroup$ – Daniel Valenzuela Mar 3 '15 at 4:13
  • $\begingroup$ @Dan I've no idea what you mean by "that"? $\endgroup$ – David Wheeler Mar 3 '15 at 4:21
  • $\begingroup$ Okay technically you did not formulate a statement, so I am sorry and take my comment back. I read the first sentence without the word "consider". $\endgroup$ – Daniel Valenzuela Mar 3 '15 at 4:26
0
$\begingroup$

The first part is already great: if every pre image is open, then in particular preimages of open sets are open. Now for the second part:

Hint: Let $f$ be continous. Then every preimage $f^{-1}(\omega)$ is open and closed, since also $f^{-1}(\Omega - \{\omega\})$ is open.

By the way this gives you a characterization of what it means to be connected for a space. I.e. $X$ is connected if and only if every map $X \to A$ into a discrete space is constant.

Edit: Let's have a closer look at those maps $f$. Any such map will be constant on connected components. Conversely you can define a continous map by assigning a constant to every connected component. Hence you get that $Maps(X,\Omega) \leftrightarrow \Omega^{\#\text{connected components}}$

$\endgroup$
1
$\begingroup$

Your argument is correct, since every subset of $\Omega$ is open, for any $f:\Omega\to X$, $f^{-1}(U)$ is open in $\Omega$ for every open $U\subset X$.

$\endgroup$

Your Answer

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