Let $A, B$ be sets and $f : A \rightarrow B$ be a function. If $f$ is injective, show that the function $g : A \rightarrow f(A)$ is bijective where $g(a) = f(a)$ for each $a \in A$.  I can barely make sense of this question, I'm beginning to lose all hope that I will ever be able to understand mathematics at this level.
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I have a function $f$ that is injective from $\mathbb R \to \mathbb  R$ but not surjective, i.e., the image $f$ is not all of $\mathbb R$. Can I make $f$ bijective by patching the codomain and removing any parts that are not in the image?
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Let $\mathcal{A}$ be the class of all sets and define the relation $R$ on $\mathcal{A}$ as: $A\space R\space B$ iff there is a bijective function $f:A \to B$. Prove that $R$ is an equivalence relation on $\mathcal{A}$.  I know I need to prove bijection by proving injection and surjection, but I don't even know where to start. Any help would be greatly appreciated.
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So here is this question I got stuck on:  Let $(a,b)$, $(c,d)$ be intervals (not sure if that's the correct term) on $\Bbb R$, so that $a&lt;b$, $c&lt;d$. Find an injective and surjective function $f:(a,b)\rightarrow(c,d)$.   Thanks in advance!   P.S. I get these question a lot. Is that some way of thinking, or some general method I should follow to make these question easier for me to solve?
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Some explanations:  A set S is  if there exists an injective function $f$ from $S$ to the natural numbers ($f:S \rightarrow \mathbb{N}$).  $\{1,2,3,4\}, \mathbb{N},\mathbb{Z}, \mathbb{Q}$ are all countable.  $\mathbb{R}$ is not countable.  The  $\mathcal P(A) $ is defined as a set of all possible subsets of A, including the empty set and the whole set.  $\mathcal P (\{\})=\{\{\}\}, \mathcal P (\mathcal P(\{\}))=\{\{\}, \{\{\}\}\} $  $\mathcal P(\{1,2\})=\{\{\}, \{1\},\{2\},\{1,2\}\}$  My question is:  Is $\mathcal P(\mathbb{N})$ countable? How would an injective function $f:S \rightarrow \mathbb{N}$ look like?
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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
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I have trouble understanding the following task:     Show that the set $X = \{f_1,\ldots ,f_6\}$ of functions $f_i : \Bbb Q\setminus \{0,1\} \to \Bbb Q\setminus \{0,1\}$ with   $x ↦ f_1(x) = x$,$x ↦ f_2(x) = 1/x$,$ x ↦ f_3(x) = 1-x,$ $x ↦ f_4(x) = 1/(1-x)$,$ x ↦ f_5(x) = (x-1)/x$, $x ↦ f_6(x) = x/(x-1)$   and the composition as the operation is a group. Is this group commutative?    I'm a little confused as I don't know what to do with all these functions. I can't apply the definitions of a group on this because of that. If the group is $(X,∘)$ how do I show that this is associative don't I need the information on how the single functions are connected or am I missing something?
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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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Is there a simple geometric proof that there exists a continuous bijective function between a square and its side?    And is there some explicit continuous function or formula $f^1(z)\mapsto (x,y)$ and $f(x,y)\mapsto z$, with $(x,y) \in [0,1]\times[0,1]$ and $z \in [0,1]$?  And is there a constructive proof that two sets of the same cardinality have a bijective function?
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Show that: Let A be a set and let $P(A)$ be the set of all subsets of $A$. Then there is no surjection $f: A→P(A)$.  Here is what I thought:  if $A=\{a,b\}$ then it has only two elements where $P(A)=\{∅,\{a\},\{b\},\{a,b\}\}$ has 4 elements. Therefore $f:A→P(A)$ cannot be surjective. But I have some problems:  1) How is it possible that any $f$ function to take $\{a\}$ from set $A$ to $\{a,b\}$? Maybe because I am thinking mainly about functions with real values like $f(x)=2x$, I find it a little bit strange that a function to take an element of a set to another set which has more elements. Is it possible?  edit: Now I thought that if $f(x)$ is $\sqrt{x}$, then $f(4)=±2$ which means it took an element from a set to a set which has 2 elements. But still I find it kind of strange to denote $f(\{a\})=\{a,b,c,...\}$  2) How can I construct a explicit proof for this question?  Regards
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Let S be a finite set.Let F be a surjective function from S to S.  How do I prove that it is injective?
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Prove that   injective function $f:X\to Y$ exists surjective function $g:Y\to X$ exists   is equivalent.$(x,y\neq \emptyset)$   My approach  First,prove left to right.  If injective function $f:X\to Y$ exists, a set A can be defined like this. $$A=\{f(x):x\in X\}$$ then it is sure that $A\subset Y$, and $|A|=|X|$  And, if we define function $g$ like $$g(y)=\begin{cases}x &amp; \text{if $y\in A$ and } f(x)=y,\\x_0&amp; \text{if }y\notin A.\end{cases}$$ ($x_0$ is a fixed element of $X$)  Then, $g$ is a surjective function, so surjective function $g:Y\to X$ exists.  Now,prove right to left.  If surjective function $g:Y\to X$ exists, $\forall x \in X$, we can select a element $y\in Y$ that $g(y)=x$, which is clear that we can pick a different $y\in Y$ for $\forall x \in X$  if we define function $f$ as $f(x)=\text{one element in Y that g(that element)=$x$}$, f is a injective function, so injective function $f:X\to Y$ exists.    But, I am not sure that my approach is right, especially on proving right to left. Please check whether my approach is right, ways to improve it,  and some other ways of proving this question.
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If we have a function $f:A \rightarrow B$, then one way to give meaning, I think, to this function, in terms of set theory, is to say, that $f$ is actually a binary relation $f=(A,B,G_f)$, where $G_f \subseteq A \times B$ is the graph of the function. Now my question is: what is $f$ if   $\bullet \ A=\emptyset, \ B\neq\emptyset$,?  $ \bullet \ B=\emptyset, \ A\neq\emptyset$ ?  $ \bullet \ B=\emptyset, \ A=\emptyset$ ?  (Another way to formulate this, I think, would be: How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset $ look like? Are they all $\emptyset$ ?)
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Let $f\colon A \to B$ and $C \subset A$. Define $f[C]=\{b∈B : b=f(a) \text{ for some } a \in C\}$. Prove or disprove each of the statements:  a) If $f$ is a bijection, then $f[A\setminus C] = B\setminus f[C]$.  Let $f$ be a bijection. since $f$ is onto then $f[A]=B$ and since $f$ is 1 to 1 then $f[A\setminus C] = B\setminus f[C]$ since every $A$ correspond to a $B$ and every $C$ correspond to some $f[C]$ so you remove that part from $B$.  Is this correct?I'm not sure if this is enough or all i need. Is there a more concise proof?   b)If $f[A\setminus C] = B\setminus f[C]$ for every $C \subset A$, then $f$ is a bijection  I'm not sure how to do this one. do I say that do I say this need to be 1 to 1 so $A$ doesn't make a $f[C]$. Any help would be great. Thanks!
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Question: Are my proofs below valid?   In both cases we are using: $f:A\to B, g: B\to C$  Notation of your type converted: $(g\circ f)(x)=g(f(x))=xfg$    If $fg$ is injective what can be said about $f$ and $g$?  If $fg$ is injective then $f$ is injective:  Proof:  $(x_1fg = x_2fg \implies x_1=x_2)$  $x_1fg=x_2fg$ is clearly true when $x_1f=x_2f$  Which requires $x_1=x_2$, hence $f$ is injective.  If $f$ were not injective, then $x_1fg\ne x_2fg$  $\blacksquare$    If $fg$ is surjective what can be said about $f$ and $g$?  If $fg$ is surjective then $g$ is surjective:  Proof:  $(\forall c\in C)(\exists a\in A) |( a(fg))=c$  This means for $g$ every element must have had been mapped to. How can this be proved Mathematically?  well $(af)g=c$ is surjective, hence $(af)=(b\in B)$ and $bg=c$  $(\forall c\in g)(\exists b\in B)|(bg=c)$  Hence $g$ is surjective.  $\blacksquare$
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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) $\forall_{A,B \subset X}: f(A \cap B) \subset f(A) \cap f(B)$ iii) $f$ is injective $ \Longleftrightarrow$ $\forall_{A,B \subset X}: f(A \cap B)=f(A) \cap f(B)$  My problem is: I know how to operate on sets, I know how to operate on sets, but I don't know how and where to start the proof,  the biggest problem in mathematics, I think.   Thanks for help!
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Is this statement true? i.e. If $f(n)$ is injective and $p_n \rightarrow L$, does $p_{f(n)} \rightarrow L$?? (I realize that it's technically only a rearrangement if $f(n)$ is bijective.)   My attempt at a proof:  Since $p_n \rightarrow L$, we have that, for all $n$ except $n \le N$, $d(p_n, L) &lt; \epsilon$.  Let $S = \{n \ | \ f(n)\le N\}$, let $n_0$ be the largest $n \in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\forall \ n &gt; n_0 \ f(n) &gt; N$ which implies that $p_{f(n)} \rightarrow L$, as required.  Is this correct? Does my proof work?
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Suppose $G$ has odd order, show the function $f:G\rightarrow G$ defined by $f(x)=x^2$ is injective. This proposition is easily provable if we assume $G$ is Abelian, but I don't know how to start this without the assumption of being Abelian.