Let $f:M\to N$ be a function. For all $Y$ subset of $N$, $f(f^{-1}(Y))$ is a subset of $Y$.  When $f$ is surjective, $f(f^{-1}(Y))$ equals $Y$. Is this true or false?  I guess because of $f(f^{-1}(Y)) = \mathrm{id}(Y)$ this statement is true. Can give some one a better explanation?
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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?
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How can one show that if $g(f(x))$ is injective and $f$ is surjective then $g$ is injective?  Here is my attempt:  $g(f(x))$ is injective, so  $$g(f(a))=g(f(b)) \iff f(a)=f(b).$$  $f$ is surjective, so $f(x)=y$.  So $y=y'$ such that $f(b)=y'$.  So $g$ is injective.  Is my proof correct?
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So as far as I understand, a function $f\colon A \to B$ is surjective if and only if for every $b\in B$ there exists $a\in A$ such that $f(a) = b$.  My question is when is this actually relevant? Couldn't you arbitrarily define the set $B$ so that any elements never "used" are removed from the set, leaving you with a surjective function?
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How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$?  I have already recognized one problem caused from this: it  follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?  Best regards and thanks, John
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Hi everyone I'd like to know if the following reasoning is correct, any suggestion would be great. Thanks.  Proposition: Let $Y$ be a set and let $f: \mathbb{N}\rightarrow Y$ be a function. Then $f[\mathbb{N}]$ is at most countable.   Proof: If $Y$ is finite the result follows, this is because$f[\mathbb{N}]\subset Y$ and any subset of a finite set  is finite and has at most the size of the set in which is contained.   We may assume  that the set $Y$ is infinite. We set $A:= \{\,n\in\mathbb{N}:f(i)\not=f(n) \,\,\text{ for  any }i&lt; n\, \}$. And we define the function $g:A \rightarrow f[\mathbb{N}]$ as the restriction of $f$ in $A$. We claim that the function $g$ is a bijective map. From the definition of $A$ we already know that the map is $1:1$. We will show that also is onto.   Let $y\in f[\mathbb{N}]$. Suppose for the sake of contradiction that there is no $n\in A$ such that $g(n)=y$. Since $y$ lies in the image of $f$, then there is some $j\in \mathbb{N}$ for which $y=f(j)$. Then either $f(j)\not= f(i)$ for any $i&lt;j$ or there exists some $i&lt;j$ such that $f(j)= f(i)$. For the former, $j$ lies in $A$ (by its definition). For the latter, the set $\{\, n\in \mathbb{N}: f(j)=f(n)\, \text{and }\,n\not=j \}$ is non empty, and hence has a minimum element. Let $m$ be its minimum element. Then $f(m)\not=f(n)$ for any $n&lt;m$ and thus $\,m\in A$ but $g(m)=f(m)= f(j)=y$. Thus any case leads to a contradiction. The result follows by reductio ad absurdum. Hence $f$ is surjective and injective, i.e., a bijective map.   Since there is a bijection between $A$ and the images of the natural numbers under $f$, both have the same cardinality. Also we already know that $A\subset \mathbb{N}$ and any set of the natural numbers is denumerable, thus $f[\mathbb{N}]$ is countable. $\Box$
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I realised I needed to show more information, which I now did:     $$f: X \rightarrow X \,\,\,\rm{and}\,\,\,\, A \subseteq X$$      Proof that: $$f(f^{-1}(A)) \subseteq A$$   This is my proof:  By defintion:   $$f^{-1}(A)=\{x \in X\mid f(x) \in A\}$$  and  $$f(A)=\{f(x) \mid x \in A\} = \{y \in X \mid \exists x \in A: y=f(x)\} \subseteq X$$  Therefore we can end the proof by a final definition:\ $$f(f^{-1}(A))=\{y \in A: \exists x \in f^{-1}(A):y=f(x)\} \subseteq A$$  Is this a legit "proof"? And is it even a proof, when i only use definitions?
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I'm trying to prove that, for a finite set $A$, if the map $f: A\rightarrow A$ is a surjection, then it's an injection.  I've looked at this post:  but the arguments there seem very round-about, proving that injectivity implies surjectivity first.  I think I've seen other proofs, too, but they all sound kind of flimsy.  They tend to go something like "If the function is surjective then no two arguments can go to the same image, since that would require some element of the range not being the image of any element."  While that's true and I see what the argument has in mind, it's no more convincing than the statement of the claim that we're trying to prove!  Without a more rigorous proof, both seem to rely on the intuition that everything has to go to one and only one thing, but we're supposed to be proving that.  But I've been trying to develop a more thorough and direct proof, and haven't been able to.  I've tried induction:     The theorem holds trivially if $|A|=0$.  Suppose the theorem holds up to $n=|X|$, and let $|A|=n+1$ and $f:A\rightarrow A$ be a surjection.  Then let $x\in A$ and $f^{-1}(x)$ be the set of all elements $y\in A$ such that $f(y)=x$...   But at this stage I want to restrict my function, but that has the disadvantage that if $f(y)$ has more than one element in it, when I pluck it out of $A$ as the domain and pluck $x$ out of $A$ as the range, then I may not be dealing with a function mapping from a set to itself and so I don't get to use the inductive hypothesis.  Any other proofs you know about?
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I am asked to consider f(x) and g(x) which are both 1-1 functions on their respective domains. I have been asked to show f ∘ g is a 1-1 function and then a followup statement of if f ∘ g is 1-1 does that mean f(x) and g(x) are both 1-1 on their domains.  The problem I am having with this question is how do I show the result? I understand when a function is 1-1 over a certain domain and I have tried composing my own functions but am unsure how I prove the result is true because I do not believe providing an example of two functions f(x) and g(x) where the first statement holds is really what I am supposed to do.  Sorry for posting again so soon, especially for such a similar and straightforward question.  Thank you.
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Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function s.t $|f(x)-f(y)|\ge \frac{1}{2}|x-y|$ for all $x,y$. Then $f$ is injective can be seen easily as $x\neq y$ implies $f(x)\neq f(y)$.  But how to check that $f$ is surjective or not ?
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Function is bijective when it is injective and surjective. Function is injective if $$(\forall x_1,x_2 \in A)f(x_1)=f(x_2)\Rightarrow x_1=x_2$$ and surjective if $$(\forall y \in B)(\exists x \in A)y=f(x)$$ Is it possible to transform a function, so that it is given in explicit form? I don't know how to manipulate with function in this form ($f(x,y)=(2x-y,x-4y)$).  Thanks for replies.
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I am trying to prove the following statement but have trouble  comprehending/going forward with some parts! Here is the statement:     If $A$ is any set, then  $|A|$ $&lt;$ $|P(A)|$   Here is what I have so far:  We need to show that there is an injection from $A$ to $P(A)$ but not a surjection.   A natural choice for an injection is the function $ f(x)$ $=$ $\{x \}$, which in plain English, takes any element $x$ (that is in $A$) and sends it to the one-element set $\{x \}$. Thus $f(x)$ is injective!  To show that there is no surjection, for the sake of contradiction, assume there is a surjection. Here is where I start to have trouble. Surjectivity means that every element of the co-domain is mapped to an element of the domain, correct? Consequently, in this particular case, we are "matching" sets (from $P(A)$) to elements (from $A$) right?  If the above is correct, my problem arises here. I am not sure how to prove that $f$ is not surjective. Unfortunately, I am easily confused by notation so please explain in English.  Thank you in advance!! :)
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For nonempty sets $A$ and $B$ and functions $ f: A \to B$ and $g: B \to A$, suppose that $ g \circ f = i_A$. Prove that if $f$ is onto, then $g$ is $1-1$.   Here is what I started, I don't think it's correct.   Suppose $f$ is onto  i.e for each element $b \in B$ there is an element of $a\in A$, $f(a)=b$. (definition of onto).   Let $y,\ z$ be any two elements of $B$.Then $y=f(a)$ and $z=f(b)$ for some $a,b \in A$.   If $g(y)=g(z)$, then $g(f(a)) = g(f(b))$,which implies that $a=b$, so $f(a)=f(b)$, so $y=z$.   Thus, whenever $g$ maps $y$ and $z$ to the same element in $A$, $y=z$, so $g$ is one-to-one.   PS: New to proof writing, any constructive criticism is much appreciated.
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I came across the functional equation $$ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 \text . $$ So far I tried plugging $ x = f ( y ) $ and got $ f ( x ) = \frac { f ( 0 ) - x ^ 2 + 1 } 2 $ which holds for every $ x $ that is equal to $ f ( y ) $ for some $ y $. I suppose that $ f ( 0 ) = 1 $ and $ f ( x ) = 1 - \frac { x ^ 2 } 2 $, which would be true if I prove that $ f $ is surjective, which I still haven't proven. Could anybody help me?
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when I apply $f^{-1}$ to both sides $f(f(x))=f(x)$ $(1)$ we get $f(x)=x$ $(2)$.  are $1$ and $2$ equivalent?
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Given an arbitrary set $A$, let $f : A \rightarrow\wp(A)$, be the function defined for all $a\in A$ by "$f(a) = \{a\}$"   Is $f$ injective?  Is $f$ surjective?    I am struggling with this question.  How am i able to identify if the function is injective? Can somebody give an example?  What confuses me is the function "f(a) = {a}"
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Question     Let the function $f:\mathbb R\to\mathbb R$,and such   $$f(xf(y)+x)=xy+f(x)$$      Find all $f(x)$   Let $x=1,y=1$,then $$f(f(1)+1)=1+f(1)$$ let $f(1)=t$,then $$f(t+1)=1+t$$  So I guess $$f(x)=x$$ is such it, and I found $f(x)=-x$ is also such it?   Can you help?  I found this simaler problem：(BMO 1997, 2000) $$f(xf(x)+f(y))=y+(f(x))^2,f:R\to R$$ this answer $f(x)=x$ or $f(x)=-x$,  the full solution can see
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Let $x=0\,\, f(x+y)=f(0+y)$ every $y \in \mathbb{Z}$ So we get $f(0)+f(y)=f(y)$ then $f(0)=0$  $0=f(0)=f(x(-x))=f(x)+f(-x)=0$ Let $f(1)=k$ for some $k \in \mathbb{Z}$ $f(n)=f(1+1+1+...+1) (n \text{ many } 1)$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$=f(1)+f(1)+...+f(1) (n \text{ many } f(1))$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$=n\cdot k$  Thus $f: \mathbb{Z} \to \mathbb{Z}$ a function such that $f(n)=nk$ I found this but I must also show that there is no other function which satisfies $f(x+y)=f(x)+f(y)$ and I am stuck can somebody help me.