To prove $f(A\cap B)=f(A)\cap f(B) \Longleftrightarrow \text{f is injective}$  Beginning:  $f(A\cap B)=\{f(x): x\in (A\cap B)\}=\{f(x): x\in A ∧ x\in B\}$ Is that correct and how can I proceed?
123601
Let $f:X\to Y$ where $X$ and $Y$ are nonempty. Prove that a sufficient and essential condition for any two subsets $A,B\subseteq X$ to fulfill $f(A\cap B)=f(A)\cap f(B)$ is that $f$ is injective. I sense there is some problem in my proof. I would be glad if you assisted me.   $Attempt:$ Let $f$ be injective and let $A,B\subseteq X$ be two subsets. If $A$ and $B$ are disjoint, then $A\cap B=\emptyset$ $\Rightarrow$ $f(A\cap B)=\emptyset$. Since $f$ is injective then there are no two elements with the same image and therefore $f(A)\cap f(B) =\emptyset =f(A\cap B)$. Now suppose $A\cap B\ne \emptyset.$ Let $ y\in f(A\cap B)$. There exists an $x\in A\cap B$ such that $f(x)=y$. Since $x \in A$ then $f(x)=y\in f(A)$ and since $x \in B$ then $f(x)=y\in f(B)$ $\Rightarrow$ $y\in f(A)\cap f(B)$. Therefore $f(A\cap B)\subseteq f(A)\cap f(B)$.   Now let $y\in f(A)\cap f(B)$. ( $f(A)\cap f(B)$ is not empty because if it were then $A$ and $B$ would have an empty intersection as well, which they don't.). Therefore $y\in f(A)$ and $y\in f(B)$ and therefore in $A$ there exists $x_1$ such that $f(x_1)=y$. The same with $B$, $f(x_2)=y$. By injectivity: $x_1=x_2 \Rightarrow x_1=x_2\in A\cap B\Rightarrow f(x_1)=y\in f(a\cap B)\Rightarrow f(A)\cap f(B)\subseteq f(A\cap B) \Rightarrow f(A\cap B)=f(A)\cap f(B).$   Necessary: Suppose it weren't necessary that $f$ is injective. Then there would exist a non-injective function $f$ such that for any two subsets $A,B\subseteq X$ we get $f(A\cap B)=f(A)\cap f(B)$. $f$ is non-injective and therefore there would be $x_1,x_2\in X$ such that $f(x_1)=f(x_2)$. Let us look at $\{x_1\},\{x_2\}\subseteq X$. $f(\{x_1\})\cap f(\{x_2\})=\{f(x_1)=f(x_2)\}\ne f(\{x_1\}\cap\{x_2\})=f(\emptyset)=\emptyset$. A contradiction.
36766
I haven't been able to do this exercise:     Let $f: A \rightarrow B$ be any function. $f^{-1}(X)$ is the inverse   image of $X$. Demonstrate that if $f$ is surjective then $X = f(f^{-1}(X))$ where $X \subseteq B$.   Since $X \subseteq B$, all the elements in $X$ belong to the codomain of $f$.  Since $f$ is surjective, it means that all elements in the codomain $B$ have some preimage in $A$. Given that $X \subseteq B$, all elements in $X$ must also have a preimage in $A$.  Have $\triangle = f^{-1}(X)$, $\triangle$ is now a set containing the preimages of the elements in $X$. Because of this, $\triangle \subseteq A$.  If we evaluate $f(\triangle)$, we...... nope, I don't know what I'm doing now.    What do you think?
55326
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, how can we show that $S = f^{-1}(f(S))$ for all subsets $S$ iff $f$ is injective?
46530
Suppose $f$ and $g$ are continuous on $[a,b]$ and that $f(a) &lt; g(a)$ but $f(b) &gt; g(b)$.   Prove that $f(x) = g(x)$ for some $x \in [a,b]$.     I did try and actually did get as far as coming up with $f-g$. Would I set $h(x)= f(x)- g(x)$ and say $h(a) &lt;x&lt; h(b)$?
21499
Statement: Let $A$, $B$, and $C$ be sets. Then $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$.   Proof: Let $(x,y) \in ( A \setminus B) \cap (A \setminus C)$, where $ x \in A$ and $y \in (B \cap C$ by definition of intersection. So, $(x,y) \in (A \setminus B \cap C)$ which show that $(A \setminus C) \cap (A \setminus C) \subseteq A \setminus (A \cup C)$. \ Now let $(x,y) \in A \setminus(B \cup C)$, where $ x \in A $. So, $ (x,y) \in (A \setminus B) \cap (A \setminus C)$ by definition of intersection, which shows that $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$. Therefore, by the Axiom Set of Equality we know that $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$ .   Am I on the right track to proving that this is a true statement or is there a better way to show this?
18951
If I have $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f$ is onto and $f\circ f\circ f = f$, how can I prove that $f$ is bijective? I know that I only have to prove that it is 1-to-1 because I'm given the fact that it's onto, but how can I use the fact that $f\circ f\circ f=f$ to prove that it is 1-to-1?  Thanks!
57175
The question I have states:   "Describe a bijection $f:A\times B\rightarrow B\times A$"  The simplicity and lack of a domain or codomain that I can look at has caused this question to throw me off.   Is describing $f(a,b)=(b,a)$ where $a\in A, b\in B$ enough to do this? To what extent do I have to prove that this is a bijection?   I can say that $(x,y)=(x',y')$ if and only if $(x=x')\ \cap \ (y=y')$ and prove injectivity and surjectivity and so on, but I am a bit confused..
52776
It would be great if someone could verify this proof.  Theorem:  $\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$  Proof:  First I prove that $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)$. Take any subset $X \subseteq A \cap B$. Then $X \in \mathcal{P}(A \cap B)$. Also, $X \subseteq A \wedge X \subseteq B$, meaning that $X \in \mathcal{P}(A) \wedge X \in \mathcal{P}(B)$. This, again, means that $X \in \mathcal{P}(A) \cap \mathcal{P}(B)$, and this proves the statement.  Then I prove that $\mathcal{P}(A) \cap \mathcal{P}(B) \subseteq \mathcal{P}(A \cap B)$. Again,  take any subset $X \subseteq A \cap B$. Using just the same arguments as above, this is also a truth.  The theorem follows.
34019
$f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$  I think I have to show that the LHS is a subset of the RHS and the RHS is a subset of the LHS, but I don't know how to do this exactly.
71674
Let $\varphi: A\to B$ and define $F:C^B\to C^A$ such that $F(f)=f\circ\varphi$. Prove the following:  if $\varphi$ is surjective then $F$ is injective.  if $\varphi$ is injective then $F$ is surjective.    For both of them I realize that I should begin with assuming the condition on $\varphi$ is true and the condition for $F$ is false and then apply the composition in some way to get a contradiction but I have no idea how to begin...
118287
I'm a little stuck with this proof. Not sure where to go. I was thinking that I could first assume that $f$ is one to one and prove that it's onto, and then assume it's onto and prove that it's one to one... but I'm not sure what to do with the knowledge that $|A|=|B|=m\in\mathbb{Z}\geq 0$.  I've tried to start by defining what it means for $f$ to be one-to-one, and going from there...  \begin{align*} f\text{ is one-to-one}&amp;\Longrightarrow f=\{(a,b)\in A\times B\mid b=f(a)\wedge f^{-1}(b)=a\}\\ &amp;\Longrightarrow \dots\\ &amp;\Longrightarrow \text{$f$ is onto} \end{align*}  but not sure where to go from there... I have a feeling that my definition is too presumptuous to begin with, since I'm trying to show that $f$ is onto... not assume that it is.
39523
If $A=\left\{1,2,3,4,5\right\}$ Then     Find Number of functions $f : A \to A$ such that $f(f(x))=f(x)$   Case $1.$ if $f$ is injective then $f(f(x))=f(x)$ $\implies$ $f(x)=x$, hence there is only one injective function which is an identity function.  Case $2.$ When $f$ is many to one function  Let us assume that $f(1)=f(2)=f(3)=f(4)=f(5)=k$ Then clearly for any $k$ from the set $A$, $f(f(x))=f(x)$  hence there are five such functions.  Are there any other possibilities?
10524
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 that $f(a)=b$. Because of $b \in (B_1 \cap B_2)$, it is true that $b \in B_1$ and $b \in B_2$, so $a \in f^{-1}(B_1)$ and $a \in f^{-1}(B_2)$.    this means $f^{-1}(B_1 \cap B_2) \subseteq f^{-1}(B_1) \cap f^{-1}(B_2)$.   is it ok?
118934
Proving $(A\cup B)\cap(B\cup C)\cap(C\cup A)=(A\cap B)\cup (A\cap C)\cup (B\cap C)$?   I managed to prove one side by saying:  Let E=LHS and F=RHS     Let $ x \in E$. Then $x \in (A\cup B) \lor \ x \in (B\cup C) \lor \ x \in (C\cup A).$ If $x\notin A$, then $x$ must be in $B$ and $C$. If $x\notin B$, then $x$ must be in $A$ and $C$. If $x \notin C$, then $x$ is in $B$ and $A$. Therefore $x \in F$ and $ E \subset F $.   I'm having trouble proving the other direction. Can I prove it in a similar way saying that let $x \in F$, if $x \notin A$, then $x$ would be in $B$ and $C$? That doesn't make sense to me. Another way I thought of solving it would be:  Let $x \in F. $ Then   if $x \in A$, then  $x \in B \cup  C$, If $x \in B$, then $x \in A \cup C$.  if $x \in C$, then $x \in B \cup A$   And then is this sufficient to show that $ x \in F, thus \ F \subset E \ and \ E=F $?
94294
Firstly, the assignment:     Let $a,b \in\mathbb{R}$ and $a &lt; b$. Furthermore let $f: [a,b] \rightarrow [a,b]$ be monotone increasing. Show that if $x:= \mathbf {sup}\{y \in [a,b] \| \ y ≤ f(y)\}$ then $f(x) = x$.   It's good on me that the assignment itself actually tells you what $x$ is so I can prove the theorem by showing that for $\forall x' \not= x: f(x) \not= x$ and I'm done. However, I have not arrived at a contradiction so far.  Any tips?
891
Does $f(f(f(x)))=x \implies  f(x)=x$? Is it necessary for it to follow? How do we prove this? Do we have to substitute some special $x$? Or is it some other consideration of its properties?  Edit: does it follow in the special case when f is surjective? Or would we wind up with the same thing?
120159
This is an exercise of an assignment I have:     Suppose $A$ and $B$ are finite sets and $f\colon A\to B$ is surjective. Is it   true that the relation “$|A| &lt; |B|$” is a sufficient condition for   claiming that $f$ is a bijection? Justify your answer.   And this is my answer:     No. In fact, if $|A| &lt; |B|$, then there should exist at least $1$ element   of $A$ that points to more than $1$ element of $B$, since all elements of $B$   must be pointed (surjective), but this is not a function,   because the same element of $A$ point to different elements of $B$.   Is my answer correct? I don't know if the question actually is asking for a proof or what, and if my answer is a proof, if correct.