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?

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    $\begingroup$ 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

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 $x\notin f^{-1}(U)$.


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