Are there two topological space $T_1$ and $T_2$ in which a continuous constant function sends an open set of the first topological space to a non-open set of the second?
Continuous constant map and topologies
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0You should try to avoid using $T_1$ and such for spaces in topology. These are used quite often to describe properties of the space. – 2011-11-25
2 Answers
Yes. $\mathbb R \owns x \mapsto c \in \mathbb R$ for any real number $c$.
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0@NateEldredge: Nevermind, it was a bad reasoning that I was applying. I try to explain. Since $f:x \in \mathbb{R} \mapsto c$ is continuous, then $f^{-1}(\{c\})$ is closed. Now $\mathbb{R}$ is closed and that's ok. But I did not immediately understand at first glance why, for example, when picking an open interval $A=(a,b)$ and considering the restriction to that interval, I got $f^{-1}(\{c\})=A$ open... when it must have been closed by continuity hypothesis. In considering restrictions one has to change the topological space, so A would be clopen and continuity holds. – 2011-11-25
For a trivial example let $X$ by $\mathbb{N}$ with the initial segment topology, and let $Y$ be $\mathbb{N}$ with the final segment topology. That is, $U\subseteq X$ is open iff $m\le n\in U$ implies that $m\in U$, and $U\subseteq Y$ is open iff $m\ge n\in U$ implies that $m\in U$. Let $f:X\to Y:n\mapsto n$. For any $A\subseteq X$, $f[A]\text{ is }\begin{cases} \text{closed,}&\text{if }A\text{ is open in }X\;;\\ \text{open,}&\text{if }A\text{ is closed in }X\;;\text{ and}\\ \text{neither open nor closed,}&\text{if }A\text{ is neither open nor closed in }X\;. \end{cases}$
Of course $X$ and $Y$ are only $T_0$, so they’re not especially nice spaces. However, we can’t actually do much better if $f$ is to be a surjection. Suppose that $f:X\to Y$ is surjective, where $X$ and $Y$ are both $T_1$. Then for each $x\in X$, $X\setminus\{x\}$ is open in $X$, but $f[X\setminus\{x\}]$ is either $Y\setminus\{f(x)\}$ or $Y$, both of which are open in $Y$.
If $f$ needn’t be surjective, there are easy examples with nice spaces. For instance, let $\mathbb{P}$ be the space of irrationals with the usual topology, and let $f:\mathbb{P}\to\mathbb{R}:x\mapsto x$. Then the only open set sent to an open set by $f$ is the empty set.