Linked Questions

0
votes
0answers
14 views

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 $$U,V\in \tau_Y, ~U\cap V = \emptyset \implies f^{-1}(U)\cap f^{-1}(V) = \emptyset.$$ ...
6
votes
6answers
4k views

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
12
votes
4answers
4k views

Is this proof correct for : Does $F(A)\cap F(B)\subseteq F(A\cap B) $ for all functions $F$?

Is this proof correct? To prove $F(A)\cap F(B)\subseteq F(A\cap B) $ for all functions $F$. Let any number $y\in F(A)\cap F(B)$. We want to show $y\in F(A\cap B).$ Therefore, $y\in F(A)$ and $...
16
votes
4answers
9k views

Show $S = f^{-1}(f(S))$ for all subsets $S$ iff $f$ is injective

Let $f: A \rightarrow B$ be a function. How can we show that for all subsets $S$ of $A$, $S \subseteq f^{-1}(f(S))$? I think this is a pretty simple problem but I'm new to this so I'm confused. Also, ...
5
votes
3answers
11k views

Proving that $C$ is a subset of $f^{-1}[f(C)]$

More homework help. Given the function $f:A \to B$. Let $C$ be a subset of $A$ and let $D$ be a subset of $B$. Prove that: $C$ is a subset of $f^{-1}[f(C)]$ So I have to show that every element ...
4
votes
3answers
12k views

Unions and Functions on Sets

Given these conditions, I seek a proof. Let $f: A \rightarrow B$ be a function, and let $X$ and $Y$ be subsets of $A$. Prove that $f(X \cup Y) = f(X) \cup f(Y)$. I can't seem to figure it out. It ...
4
votes
1answer
6k views

Union of preimages and preimage of union

Is it possible to have a map $f:X\to Y$ from a topological space $X$ to a set $Y$ and some subsets of $Y$ namely $U_i,i\in I$ such that $\bigcup_{i\in I} f^{-1}(U_i)$ is not equal to $f^{-1}\left(\...
7
votes
3answers
4k views

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 $f^{-1}(f(A))=A$?
7
votes
4answers
10k views

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 ...
4
votes
4answers
1k views

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 ...
6
votes
4answers
1k views

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 $C$...
4
votes
3answers
7k views

Maps - question about $f(A \cup B)=f(A) \cup f(B)$ and $ f(A \cap B)=f(A) \cap f(B)$

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(...
5
votes
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)\...
5
votes
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 $f$ ...
7
votes
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 ...

15 30 50 per page