Let $f: A \to B$ be some function, and we declare $g:P(A) \to P(B)$ by $g(Y) = f^{-1}[Y]$. I need to prove that  $f$ is surjective iff $g$ is injective.   So I started with the first direction, I assume $f$ is surjective and want to prove that $g$ is injective. So I assume $g(X) = g(Y)$ for some subgroups of $B$, which means that $f^{-1}[X] = f^{-1}[Y]$. But I get stuck. I don't see how the fact that $f$ is surjective helps me here,  since the only thing I can conclude is that  $f^{-1}[X]$ and  $f^{-1}[Y]$ are not empty.   I am also having trouble with the second direction.  Moreover, does $f$ being surjective imply $g$ is surjective?  Help would be appreciated.
121330
So the idea that I am having is, saying that we let f be surjective, and we would prove that $f^{-1}$ is injective. What is confusing me is how I would word my proof and how I would represent it. I am starting it off with Let X and Y be non-empty sets, and let $f: X\rightarrow Y$ be a function. Now I'm stuck.
71829
(1) Prove or disprove: For any sets $X,Y,Z$ and any maps $f:X \to Y$ and $g:Y \to Z$, if $f$ is injective and $g$ is surjective, then $g \circ f$ is surjective.  So i proved previously that $f$ is injective if $g \circ f$ is injective, but then i also proved that if $g \circ f$ is surjective then $g$ is surjective. then this question appears, so how do i prove it that it is true or false? I'm saying it's true, because $g \circ f$ is bijective, so how do i go around proving it? Or my idea is already wrong? I need some help thanks.  (2) Show that (it is true in general that) for any sets $A,B$, one has $P(A) \cup P(B) \subseteq P(A \cup B)$.  I know that power set has exactly all the subsets of $A$ for $P(A)$. but I dont know how to start.
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I am working on a school assignment and have been stuck on this question for some time.   Let $f : G \rightarrow G_1$ be a surjective homomorphism (also called epimorphism) from $G$ to another group $G_1$. Prove that $f(Z(G)) \subseteq Z(G_1)$  I know we start by assuming we have an element in $f(Z(G))$ then showing it is also in $Z(G_1)$. Also, $Z(G)$ is the center of $G$, so $Z(G)$ is all the elements of $G$ that commute with all elements in $G$. But how do we know that if we map the elements of $Z(G)$ to $G_1$, then their image will also commute with all the elements in $G_1$? Does it have something to do with using commutativity with the homomorphism? i.e. $f(gz) = f(g)f(z) = f(zg) = f(z)f(g)$.
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I don't follow understand what this theorem is saying: "Every function $f: A \to{\cal P}(A)$ is not surjective."   Is this saying so for example let A = {1,2,3} then the power set of A is {},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3} Is this saying that there is no surjective function that takes every element in each of its power sets to atleast 1 element in A?  Is my understanding correct?
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Specifically, $f = g$ on $S$, where $S$ is a dense subset of $X$. Is it true that $f = g$ on all of $X$? I'm quite sure that this holds true in the case where $X = \mathbb{R}$, but I don't think it holds true in general. Is this true? If so, could you provide a counterexample where $f \neq g$ on $X$?
56780
I'm doing exercise on discrete mathematics and I'm stuck with question:     If $f:Y\to Z$ is an invertible function, and $g:X\to Y$ is an invertible function, then the inverse of the composition $(f \circ\ g)$ is given by $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$.   I've no idea how to prove this, please help me by give me some reference or hint to its solution.
104866
I keep going around in circles with this problem:  Given the functions $f:X\to Y$ and $g:Y\to Z$, if $g\circ f:X\to Z$ is bijective, then prove $f$ must be one-to-one and $g$ must be onto. I was able to prove this without difficulty. However the second part of the problem has me stumped:  a) Find functions $f,g$ where $g\circ f$ is bijective but $f$ fails to be onto.  b) Find functions $f,g$ where $g\circ f$ is bijective but $g$ fails to be one-to-one.  My attempt for (a): Let $f:\mathbb{R}\to \mathbb{R}^+$ be defined by $f(x)=e^x$ and $g:\mathbb{R}^+\to \mathbb{R}$ be defined by $g(x)=\ln(x)$. Then $g\circ f(x)=g(f(x))=g(e^x)=\ln(e^x)=x$ which is bijective, $g$ is onto and $f$ fails to be onto... or so I thought. Upon closer inspection $f$ $\underline{is}$ onto for the given codomain $\mathbb{R}^+$.  And now I'm stuck in a rut. Anyone out there have any hints?
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Problem:  Let T be a nonempty set and A(T) the set of all permuations of T. Show that A(T) is a group under the operation of composition of functions.  Permutation of the set T is the bijective function $f: T \rightarrow T $  For A(T) to be a group it must be closed, associate have an identity for every element and an inverse.  Since f is  bijection the composition of two bijective functions is itself bijective so A(T) is closed. Every bijective function has an inverse, and composition of functions is associative. What about the identity? Are we to take $ A(T) = S_n $ here, which would be assuming that $T$={$1, ..., n$}? In which case $ I = \begin{bmatrix}1&amp;2&amp;...&amp;n\\1&amp;2&amp;...&amp;n\end{bmatrix} $  Not sure what I'm supposed to be showing since $T$ and A(T) are arbitrarly specified here.
101779
Let $f \colon M \to N$ a function. Show that the following affirmations are equivalent: (a) $f$ is injective (b) every two subsets $M_1$ and $M_2$ of $M$, $f(M_1 \cap M_2)=f(M_1) \cap f(M2)$  (a) $\implies$ (b): Let $y \in f(M_1 \cap M_2)$ $\implies$ exists $x$ an element from $M_1 \cap M_2$ such that $f(x)=y$ $\implies$ exists $x \in M_1$ such that $f(x)=y$ and exists $x \in M_2$ such that $f(x)=y$ $\implies y \in f(M1) \cap f(M2)$. So, $f(M_1 \cap M_2)$ is a subset of $f(M_1) \cap f(M_2)$. Let $y \in f(M_1) \cap f(M_2)$. How can I prove, using the injectivity of the function $f$,that $y \in f(M_1 \cap M_2)$?  (b)$\implies$ (a) $f(M_1 \cap M_2)=f(M_1) \cap f(M_2)$ means that every $y$ from $f(M_1 \cap M_2)$ belongs to $f(M_1) \cap f(M_2)$ and  every $y$ from $f(M_1) \cap  f(M_2)$ belongs to $f(M_1 \cap M_2)$. How can I prove that $f$ is injective?
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If $f$ is continuous from $ [0,1]$ to itself and $f(0)=0$, $f(1)=1$ and $f(f(x))=x$ then show that $f(x)=x$ for every $ x $ in the domain of function.  I'd like to prove this statement, my idea goes like below:  Suppose that the converse of the statement holds. then without loss of generality, assume that there exists some $c$ in $[0,1]$ such that $d=f(c)&gt;c$. By symmetry, $f(d)=f(f(c))&lt;f(c)=d$. And $f$ is bijective because $f$ has an inverse function in this domain. So $f$ should be strictly increasing function. But on $[c,d]$, $f$ is decreasing. This is a contradiction.  So it should be that $f(x)=x$ on $[0,1]$.  Is this argument right? And I'd like to know another argument which prove this statement. Thank you for your answer in advance :)
20115
Is it true that for each set $M$ a given injective function $f: M \rightarrow M$ is surjective, too?  Can someone explain why it is true or not and give an example?
43379
I need help solving the next question, I think that it's actually a refutation:     let $f,g: \mathbb{N} \rightarrow \mathbb{N}$  so that $\forall n ∈ \mathbb{N} : f(n) = g(2n)$   Prove that if $f$ is surjective $g$ is not injective
78422
From wikipedia I obtain the following definition of an injective function :     Let $f$ be a function whose domain is a set $A$. The function $f$ is injective if for all $a$ and $b$ in $A$, if $f(a) = f(b)$, then $a = b$; that is, $f(a) = f(b)$ implies $a = b$.    From this I conclude that a function $f$ is injective if the below statement is true for all $a,b \in A$:  $$f(a)=f(b) \implies a=b$$  My question is: Can I re-formulate the above statement as $f(a)=f(b) \iff a=b$ ?
61125
How can I formalize that the function $$f:\mathcal{P}(A)\to(A\to \{0,1\})$$ is bijective? It seems clear to me but I don't know how to write a proof. Thanks in advance.
1509
Suppose $X$ is a set and $f : X \to X$ is a function.  If $X$ is a finite set, prove that $f$ is injective if and only if $f$ is surjective. Show that when $X$ is an infinite set, the statement is not true in general.   For that, let $X = \mathbb{R}$ and find a function $f: \mathbb{R} \to \mathbb{R}$ which is injective but not surjective, and a function $g: \mathbb{R} \to \mathbb{R}$ which is surjective but not injective  My attempt at the first part   Suppose that the function is surjective but not injective. Let $a,b \in X$ such that $f(a)=f(b)$ but $a \not =b$. Now since this is a finite set mapping to itself then there are not enough elements in $X$ to map to $X$ since two elements were used to map to one element, thus the function can't be surjective. Therefore if the function is surjective it is also injective.   Now suppose that the function is injective but not surjective. Let $a \in X$ such that $f(b)\not =a$ for all $b \in X$. Since the function was assusmed to be injective each $a \in X$ must map to an element of $X$, thus must have two elemetns in $X$ mapping to the same element in $X$. Hence the function can't be injective, therefore if the function is injective it must be surjective.   Would an exponetial function work for injective but not surjective since can't map to negative numbers. And an example of a function that is surjective but not injective could be a log function since none of the negative numbers get mapped to anything?   Also I'm having trouble seeing why it has to be finite for the first part to work. I understand it has something to do with the pigeonhole principle but its not that clear to me.
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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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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?