# Linked Questions

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### Intersection of images is empty implies intersection of preimages is empty. [duplicate]

Let $f:(X,\tau_X) \rightarrow (Y,\tau_Y)$ be a continuous function between topological spaces. Can you show that $$U,V\in \tau_Y, ~U\cap V = \emptyset \implies f^{-1}(U)\cap f^{-1}(V) = \emptyset.$$ ...
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### Do we have always $f(A \cap B) = f(A) \cap f(B)$? [closed]

Suppose $A$ and $B$ are subsets of a topological space and $f$ is any function from $X$ to another topological space $Y$. Do we have always $f(A \cap B) = f(A) \cap f(B)$? Thanks in advance
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### Is $f^{-1}(f(A))=A$ always true?

If we have a function $f:X\rightarrow Y$ where $A\subset X$, is it true to say that $f^{-1}(f(A))=A$?
4answers
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### Prove $F(F^{-1}(B)) = B$ for onto function

Suppose that $f:X \to Y$ is an onto function. Prove that for all subsets $B$ subset of $Y$, $f(f^{-1}(B)) = B$. I don't know how to do this if the function is not also one to one, which it is not. Any ...
4answers
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### Why $f^{-1}(f(A)) \not= A$ [duplicate]

Let $A$ be a subset of the domain of a function $f$. Why $f^{-1}(f(A)) \not= A$. I was not able to find a function $f$ which satisfies the above equation. Can you give an example or hint. I was asking ...
4answers
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### Proving $f(C) \setminus f(D) \subseteq f(C \setminus D)$ and disproving equality

Let $f: A\longrightarrow B$ be a function. 1)Prove that for any two sets, $C,D\subseteq A$ , we have $f(C) \setminus f(D)\subseteq f(C\setminus D)$. 2)Give an example of a function $f$, and sets $C$...
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I am struggling to prove this map statement on sets. The statement is: Let $f:X \rightarrow Y$ be a map. i) $\forall_{A,B \subset X}: f(A \cup B)=f(A) \cup f(B)$ ii) $\forall_{A,B \subset X}: f(... 4answers 1k views ###$f(A \cap B)\subset f(A)\cap f(B)$, and otherwise? I got a serious doubt ahead the question Be$f:X\longrightarrow Y$a function. If$A,B\subset X$, show that$f(A \cap B)\subset f(A)\cap f(B)$I did as follows$$\forall\;y\in f(A\cap B)\... 1answer 5k views ### Show that$f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$Show that$f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B)$but not necessarily$f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$. Let$S=A\cup B$I know that$f^{-1}(S)=\{x:f(x)\in S\}$assuming that that$f$... 2answers 3k views ### how to prove$f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$I am given this equation:$f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$I want to prove it: what i did is I take any$a \in f^{-1}(B_1 \cap B_2)$, then there is$b \in (B_1 \cap B_2)\$ so ...

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