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Given two sets $x = \{ a_1, a_2, a_3, a_4 \}$ and $y = \{ \emptyset, x, \{ a_1, a_2, a_3\}, \{ a_3 \}, \{ a_3, a_4 \} \}$, where $y$ is a topology defined on $x$.

How could we construct a continuous function $f:x\rightarrow y$ without $f(a_i) = a_i$, i.e. fix points ?

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  • $\begingroup$ Continuous with respect to which topologies? $\endgroup$ – J.-E. Pin Feb 3 '16 at 22:56
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    $\begingroup$ The constant function $a_3$, no? I am assuming that the second list provides the open sets. $\endgroup$ – lulu Feb 3 '16 at 22:56
  • $\begingroup$ how define continuity? $\endgroup$ – AsdrubalBeltran Feb 3 '16 at 22:58
  • $\begingroup$ Or were you asking for a continuous function with no fixed points? The way I read it at first, I thought you just wanted a continuous function other than the identity (which isn't continuous). Either way, I think your question should be edited...it really isn't clear. $\endgroup$ – lulu Feb 3 '16 at 23:12
  • $\begingroup$ Is $y$ the topology on $x$? And do you want $f\colon x\to x$? $\endgroup$ – BrianO Feb 3 '16 at 23:21
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Assuming that the second list, $y$ is the list of open sets and that what's wanted is a continuous function $f: x \to x$ with no fixed points, then consider:

$$\{a_1,a_2,a_3,a_4\}\to\{a_2,a_1,a_1,a_1\}$$

(i.e. $f(a_2)=f(a_3)=f(a_4)=a_1,\;f(a_1)=a_2$)

Then we check:$$f^{-1}(x)=x,\; f^{-1}(\emptyset)=\emptyset,\;f^{-1}(\{a_1,a_2,a_3\})=x,\;f^{-1}(a_3)=\emptyset,\; f^{-1}(\{a_3,a_4\})=\emptyset$$

so in each case, the inverse image of an open set is open. By inspection, this function has no fixed points.

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  • $\begingroup$ Does $\{ a_1, a_2, a_3, a_4 \}\rightarrow \{ a_2, a_1, a_1, a_1 \}$ mean that for function $f$, we have $a_1$ maps to $a_2$, and each of $a_2, a_3, a_4$ maps to $a_1$, and nothing maps to $a_3$ or $a_4$ ? $\endgroup$ – thm Feb 3 '16 at 23:37
  • $\begingroup$ Exactly. Is this what you were after? $\endgroup$ – lulu Feb 3 '16 at 23:39
  • $\begingroup$ How could we infer that e.g. $f^{-1}(\{a_1, a_2, a_3\}) = x$ ? Since f it not a bijective mapping, there is no inverse for $f^{-1}(a_1)$, although we can have $f^{-1}(a_2) = a_1$ and $f^{-1}(a_3) = \emptyset$ $\endgroup$ – thm Feb 4 '16 at 0:03
  • $\begingroup$ If $F$ is a map from $X$ to $Y$ and $S$ is a subset of $Y$, then $F^{-1}(S)$ means the set of all $x\in X$ such that $F(x)\in X$. So, for my function, $f^{-1}(\{a_1,a_2,a_3\})$ means the set of all values wich map into that set. $\endgroup$ – lulu Feb 4 '16 at 6:10

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