Question: Let $X=\lbrace 1,2,3,4,5 \rbrace$ with topology $\lbrace \emptyset,X,\lbrace 1 \rbrace,\lbrace 3,4 \rbrace,\lbrace 1,3,4 \rbrace\rbrace,$ and let $Y=\lbrace A,B\rbrace$ with topology $\lbrace \emptyset,Y,\lbrace A\rbrace\rbrace.$ Find all continuous functions from $X\to Y$.

I know that given any open subset U in Y, f^{-1}(U) must be open in X. Now, how do I list and find these functions for the above spaces?


You need $f^{-1} V \in \tau_X$ for all $V \in \tau_Y$. The only member of $\tau_Y$ that is relevant is $\{A\}$. To be continuous, we must have $f^{-1}\{A\} \in \tau_X$. There are only $5$ possibilities, and $f$ must take the value $B$ on the complement.

The functions are straightforward to list. For example, if $f^{-1}\{A\} = \emptyset$, then $f(x) = B$ for all $x$. Another, if $f^{-1}\{A\} = \{1\}$, then $f(1) = A$ and $f(x) = B$ for all $x \neq 1$.

  • $\begingroup$ Is it possible that the function maps everything in X to {A}? $\endgroup$ – Julius Jackson Oct 25 '13 at 3:23
  • $\begingroup$ @JuliusJackson: Yes, if you set $f^{-1} \{A\} = X$, then $f(x) = A$ for all $x$. $\endgroup$ – copper.hat Oct 25 '13 at 3:25
  • $\begingroup$ How can I tell the total possibilities for the function f? $\endgroup$ – Julius Jackson Oct 25 '13 at 3:49
  • $\begingroup$ To be continuous, you must have $f^{-1}V$ be open for all open $V$. Since there are only 3 possibilities for $V$ it is easy to check. $f^{-1} \emptyset$ is always $\emptyset$, so there is nothing you can do about that and $f^{-1} Y $ is always $Y$, so the only open set that really matters is $\{A\}$. Then $f^{-1} \{A\}$ must be one of the 5 sets in the $X$ topology, and once you select one of the 5, you have completely defined $f$ (since the only other value it can take is $B$). Hence there are exactly five continuous functions. $\endgroup$ – copper.hat Oct 25 '13 at 4:49

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.