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Let $f :X \rightarrow Y$ be a function and suppose that $\mathfrak T_Y$ is a topology on $Y$. Let $\mathfrak T_X = \{f^{-1} (U) : U \in \mathfrak T_Y\}$ then $\mathfrak T_X$ is a topology on X.

$f^{-1}(U)$ refers to the pre-image of a set

determine if this is true or false and then prove or give a counterexample. I think it is a false statement because this conjecture doesn't say anything about the function being continuous and then we cannot find $\{f^{-1} (\phi )=\phi$ and $\{f^{-1} (Y )=X $

IS IT CORRECT ,ANY HELP PLEASE

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  • $\begingroup$ $f^{-1}(X)$ makes no sense: $X$ is the domain but $f^{-1}(U)$ only makes sense when $U$ is a subset of the range $Y$. $\endgroup$ – Lee Mosher Apr 27 '17 at 23:28
  • $\begingroup$ Before even talking about continuity of $f$ you would first have to define what topologies on $X$ and $Y$ you're using. As it turns out, if the conclusion is true, then using $\mathfrak{T}_Y$ as the topology on $Y$ and $\mathfrak{T}_X$ as the topology on $X$, $f$ is indeed continuous. $\endgroup$ – Daniel Schepler Apr 27 '17 at 23:30
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    $\begingroup$ Hint: Just by basic set theory, $f^{-1}(\emptyset) = \emptyset$, $f^{-1}(Y) = X$, $f^{-1}(\bigcup_{i\in I} V_i) = \bigcup_{i\in I} f^{-1}(V_i)$, $f^{-1} (V_1 \cap V_2) = f^{-1}(V_1) \cap f^{-1}(V_2)$. $\endgroup$ – Daniel Schepler Apr 27 '17 at 23:31
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    $\begingroup$ The question is asking you to determine whether or not $\mathfrak{T}_X$ is a topology for $X$, this can be answered by appealing to the definition of a topology and using inverse images of corresponding sets on $Y$ via $f$, as Daniel Schepler pointed out nicely. $\endgroup$ – Justin Benfield Apr 28 '17 at 0:15
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(1). For any non-empty finite $F\subset T_X$ let $F=\{f^{-1}U: U\in G\}$ where $G$ is a non-empty finite subset of $T_Y.$ We have $\cap G\in T_Y,$ so $f^{-1}(\cap G)\in T_X.$

For any $x\in X$ we have $x\in \cap F \iff$ $\forall U\in G\;(x\in f^{-1}U)\iff$ $ \forall U\in G\; (f(x)\in U)\iff$ $ f(x)\in \cap G\iff$ $ x\in f^{-1}(\cap G).$

So $\cap F=f^{-1}(\cap G)\in T_X.$

(2). For any $F\subset T_X$ let $F=\{f^{-1}U: U\in G\}$ where $G\subset T_Y.$ We have $\cup G\in T_Y,$ so $f^{-1}(\cup G)\in T_X.$

For any $x\in X$ we have $x\in \cup F\iff$ $ \exists U\in G\;(x\in f^{-1}U)\iff$ $ \exists U\in G\;(f(x)\in U)\iff$ $ f(x)\in \cup G \iff$ $ x\in f^{-1}(\cup G).$

So $\cup F=f^{-1}(\cup G)\in T_X.$

(3). We have $\phi=f^{-1}\phi\in T_X$ and $X=f^{-1}Y\in T_X.$

From (1),2),(3), we see that $T_X$ is indeed a topology on $X.$ It is the weakest topology on $X$ for which $f:X\to Y$ is continuous.

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