1
$\begingroup$

Definition: Let $X, \tau$ and $Y, \tau'$ be topological spaces. Then a function $f$ from $X$ to $Y$ is said to be continuous if given any open subset $U$ of $Y$ then $f^{-1}(U)$ is an open subset of $X$.

Okay, so one of the examples states that if $X$ is any space with the trivial topology and $f$ is any function from $X$ onto a space $Y$, then $f$ is continuous if and only if $Y$ has the trivial topology. For if $Y$ has the trivial topology, then $Y$ and $\emptyset$ are the only open subsets of $Y$; hence $f^{-1}(U)$ is open in $X$ (being either $\emptyset$ for $U = \emptyset$ or $X$ for $U = Y$). for any open subset $U$ of $Y$. if $Y$ does not have the trivial topology, then there is an open subset $U$ of $Y$ which is neither $Y$ nor $\emptyset$. Then $f^{-1}(U)$ is neither $X$ nor $\emptyset$, and hence is not an open subset of $X$. Therefore $f$ could not be continuous.

So, I don't agree with this. Now, it states that $X$ is onto a space $Y$, but nowhere does it state that $f$ is one-to-one. Or, would that only be applied for multiple elements in $X$ mapping to one element in $Y$? My argument is that we could take the open set $U$ in $Y$ and map it back to $X$ just like $Y$ maps to $X$. This would mean that $f^{-1}$ is not one-to-one, but shows that $f$ would indeed be open by definition. Am I wrong? I believe I'm wrong and the implication is that it wouldn't be possible for $f^{-1}$ to not be one-to-one. Is there an implication I'm missing?

$\endgroup$
  • 1
    $\begingroup$ The argument that you don't like is perfectly correct. If $Y$ has a nontrivial open set $U$, then since $f$ is onto, $f^{-1}(U)$ cannot be all of $X$. Note that $f^{-1}(U)$ means the set of points of $X$ mapped to elements of $U$. The inverse function of $f$ need not exist (because nothing in the argument says that $f$ is one to one). I believe, because of your use of $f^{-1}$ "naked" that you are confusing the inverse image of a set under $f$ with some sort of inverse function. $\endgroup$ – André Nicolas Apr 3 '14 at 7:22
  • $\begingroup$ Well, so the inverse image of $Y$ is $X$ and the inverse image of $\emptyset$ is $\emptyset$. How come the inverse image of $U$ can't be $X$ as well? $\endgroup$ – David Apr 3 '14 at 7:30
  • $\begingroup$ Actually, I suppose the only way for that to be possible is for $U$ to be $ Y $ $\endgroup$ – David Apr 3 '14 at 7:37
  • $\begingroup$ In my comment, I specified that $U$ is a non-trivial open subset of $Y$. In particular there is a $b\not\in U$. Since $f$ is onto, $b=f(a)$ for some $a\in X$, and $a\not\in f^{-1}(U)$, so $f^{-1}(U)$ is not all of $X$. $\endgroup$ – André Nicolas Apr 3 '14 at 7:43
2
$\begingroup$

Your argument is wrong. Note that $f^{-1}(U)$ cannot be all of $X$, as we have some $y\in Y\setminus U$ and since $f$ is onto we have some $x\in X$ such that $f(x)=y$, so $x\notin f^{-1}(U)$.

$\endgroup$

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.