Let $a,b\in \mathbb R$ with $a\neq 0$. We shall denote by $f_{a,b}$ the map from $\mathbb R$ to $\mathbb R$ such that $f(x)=ax+b$ is a bijection.  How can I show that $f_{a,b}$ is a bijection? I understand for something to be bijective it must be injective and surjective, so do I need to show it has those properties instead?
121202
$G(x) = ax + b$ where $a$ is not equal to $0$ and $b$ are real numbers.  Prove $G$ is a one-to-one correspondence.  I understand that for every $a$ there is a corresponding $b$-value that does not repeat; however, I do not understand how I can prove this.
45353
My intent is to show that a composition of bijections is also a bijection by showing the existence of an inverse.  But my approach requires the associativity of function composition.  Let $f:  X \rightarrow Y, g:  Y \rightarrow Z, h:  Z \rightarrow W$ be functions.  $((f \circ g) \circ h)(x) = h((f \circ g)(x)) = h(g(f(x)))$, and $(f \circ (g \circ h))(x) = (g \circ h)(f(x)) = h(g(f(x)))$.  However, I am having problems in justifying that the two compositions, $(f \circ g) \circ h$ and $f \circ (g \circ h)$, have the same domain and range.  When I consulted ProofWiki, whose link is at the bottom, I got even more confused.  Specifically, for $(f \circ g) \circ h = f \circ (g \circ h)$ to be defined, ProofWiki requires that dom$g =$ codom$f$ and dom$h =$ codom$g$.  First of all, I think that it should be dom$g =$ range$f$ ....  Moreover, as you can see in the example below, you actually have to adjust domains and ranges of $f, g, h$ for the requirement to hold true.  Let $f: \mathbb R \rightarrow \mathbb R$ be $f(x) = 2x$, $g: \mathbb R^+ \rightarrow \mathbb R$ be $g(y) = ln(y)$, $h: \mathbb R \rightarrow \mathbb R$ be $h(z) = z - 10$.   Then $((f \circ g) \circ h)(x) = ln(2x) - 10 = (f \circ (g \circ h))(x)$, with dom$((f \circ g) \circ h) = \mathbb R^+$ = dom$(f \circ (g \circ h))$.  As a result, we need to set dom$f = \mathbb R^+$, range$f = \mathbb R^+$; dom$g$, range$g$, dom$h$, and range$h$ remain the same.  Am I allowed to do that?  This adjustment implies that when we say dom$f = X$, $f$ must be defined for all elements in $X$, but $X$ may not be the entire set of elements for which $f$ is defined.
1896
This seems to be a common result. I've been trying to follow the bijective proof of it, which can be found easily online, but the explanations go over my head. It would be wonderful if you could give me an understandable explanation of the proof and let me know how I'd go about finding such a bijection.
120311
This proof seems odd to me. I have come to the conclusion that I will use induction. I would like to see a smoother way or just some improvements on my technique.     Let $f:A\rightarrow B$ be a function between two finite sets of equal cardinality. Show that $f$ is surjective if and only if it  is injective.   To start, I will show that a surjection implies an injection using induction. I will dismiss the cases that both sets are empty or contain one element as being trivial (essentially vacuously true).   Assume $|A| \geq 2$, $|B| \geq 2$, $|A| = n = |B|$, and $f:A \rightarrow B$ is a surjection. For the base case, let $n = 2$.   There are two elements in both $A$ and $B$. Due to surjection, every element $b \in B$ must be mapped to, through $f$, by at least one element $a \in A$. If each of the two elements in $B$ were mapped to by the same element in $A$, the definition of function would be violated. Therefore, they are mapped to by unique elements in $A$. Thus, for $f(p), f(q) \in B$, if $f(p) = f(q)$, it must be true that $p = q$ so $f$ is injective.  Now assume that the surjection implies an injection for $n \geq 2$. We must show this to be true for $|A| = n + 1 = |B|$. Since it is true for $|A| = n = |B|$, the $n + 1$ case represents the addition of one new element to both $A$ and $B$. The new element in $B$ cannot be mapped to any other element in $A$ except for the new one. If mapped to by an old one, the definition of function would be violated. It must be mapped to by something since $f$ is surjective, hence it must be the new element. Finally, the new element in $A$ cannot be mapped to an old element in $B$ because it is unique and the previous $B$ was shown to be injective.  $$\blacksquare$$  This is a very wordy and awkward proof in my opinion. I have been out of proofs for a long time. I would like to see one that is more clear or seek validation if there isn't. I know that I have only completed half of the proof and have yet to go the other way.
65235
I am stuck on this problem:  Define $f: N\times N \rightarrow N$ by   $f(i,j)=\frac{(i+j-1)(i+j-2)}{2}+i$.   How do you prove that $f$ is a bijection thus $N\times N$ and $N$ are numerically equivalent?  Work: First I tried to prove that $f$ is injective by setting $f(i,j)=f(x,y)$. However, from here I am stuck and I cannot reduce it properly. My work really does not get anywhere significant and its really messy.
2000
Let $A,B$ be two sets. The  states that if there is an injection $f\colon A\to B$ and an injection $g\colon B\to A$, then there exists a bijection $h\colon A\to B$.  I was wondering whether the following statements are true (maybe by using the AC if necessary):   Suppose $f \colon A\to B$ and $g\colon B\to A$ are both surjective, does this imply that there is a bijection between $A$ and $B$. Suppose either $f\colon A\to B$ or $g\colon  A\to B$ is surjective and the other one injective, does this imply that there is a bijection between $A$ and $B$.
107131
If we set $A$ and $B$. Assuming that there exist an injective map from $A$ to $B$ and an injective map from $B$ to $A$. How to prove that there is a bijection between $A$ and $B$.
109540
Here are two sets which I need to define a bijection function in order to show they are equivalent: $$ (A^B)^C \triangleq \{f:C\rightarrow \{g:B\rightarrow A\} \} $$ $$ (A^{B \times C}) \triangleq \{ h:B \times C \rightarrow A\} $$ where $A, B, C$ are sets, and $f, g, h$ represent functions.  I am trying to define a function $F : C \rightarrow A$ , but still struggling with its formal description, and how to use functions like $g$ above.  What is the right way of showing that such bijection $F$ exists?
115424
Prove the following are surjective, or disprove with a counter-example:   $f\colon \mathbb{Q} \to \mathbb{Q}$, $f(x) = 1 + 2x$. $f\colon  \mathbb{Z} \to \mathbb{N}\cup\{0\}$, $f(x) = |1 - x|$. $f\colon \mathbb{Q} \to \mathbb{Q}$, $f(x) = 4 - 2x^3$. $f\colon \mathbb{Z}^2 \to \mathbb{Z}$, $f(x, y) = x + y$.   Please show me the most effective method to lay out such proofs.  These come from a manual on Set Theory, which I am trying to reach to myself. Please be understanding!
78192
I already asked a question here  I am trying to prove that $\mathbb{R}^2 \cong  \{(x,y)\in \mathbb{R}^2|x^2+y^2&lt;1\}$ The function that apparently solves this is $$f(z)=\frac{z}{||z||+1}$$ for $z\in \mathbb{R}^2$.  In order to show this, I need to prove that $f$ is a continuous bijection. Continuity is obvious, but I see no reason why this is a bijection-- I also see no reason why it should have a continuous inverse. Can someone explain this to me/point me in the right direction? Thanks
74204
Let $f:\mathbb R \to \mathbb R$ be a continuous function such that $|f(x)-f(y)|≥|x-y| ,\forall x,y \in \mathbb R  $ , then how do we prove that $f$ is surjective ?
4030
Maybe I am not good at looking for the right questions but I haven't seen this task anywhere so I hope it is no duplicate.  I have to prove the following statement:     Let $V$ be a finite dimensional vector space and $f:V \rightarrow V$ a linear map. Show that $f$ is injective $\Longleftrightarrow$ $f$ is surjective.   The following is already known:  $(i)$ $\ker f$ and $\def\Im{\operatorname{Im}}\Im f$ are linear subspaces $(ii)$ There exists an isomorphism $V/\ker f\rightarrow \Im f$ $(iii)$ If $U\subset V$ is a linear subspace then $\dim V/U=\dim V-\dim U$ $(iv)$ $f$ is injective $\Longleftrightarrow$ $\ker f=\{0\}$  So my approach is the following:  $"\Longrightarrow "$ Let $f:V\rightarrow V$ be injective. Then due to $(iv)$ $\dim\ker f=0$. Hence due to $(ii)$ and $(iii)$ $\dim V/\ker f=\dim V=\dim\Im f$. Since $\dim V=\dim\Im f$ it implies that $f$ is surjective.  $"\Longleftarrow "$ Let $f$ be surjective. Then $\dim\Im f=\dim V$. Additionally, because of $(ii)$ it is $\dim\Im f=\dim V/\ker f$. Thus $\dim V=\dim V/\ker f=\dim V-\dim\ker f$. This means that $\dim\ker f=0$ and hence $\ker f=\{0\}$ and thus $f:V\rightarrow V$ is injective.  For me it all makes sense but maybe I missed something.  Thank you in advance.
41993
Let $f:\mathbb{R}\to \mathbb{R}$ is a function such that for all real $x$ and $y$, $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$, then prove that $f$ must be one of the two following functions:   $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=0$ for all real $x$   OR   $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x$ for all real $x$   I got to the point where putting the two equations together, you get $f(x+y)f(x)= f(xy) + f(x)^2$ and plugging in $f(x)=x$ checks with it. So am I going in the right direction or am I just doing some guess work? Is there a more elegant way of doing it?   Thanks
101837
A continuous bijection $f:\mathbb{R}\to \mathbb{R}$ is an homeomorphism. With the usual metric structure.   I always heard that this fact is true, but anyone shows to me a proof, and I can't prove it. I was tried using the definition of continuity but it is impossible to me conclude that. I tried using the fact that $f$ bijective has an inverse $f^{-1}$ and the fact that $ff^{-1}$ and $f^{-1}f$ are continuous identity, but I can't follow.
75790
Is there a bijection from a finite (closed) segment of the real line to $\mathbb{R}$? For example, is there a bijection from $[0,1]$ to $\Bbb{R}$?  If so, is there a straightforward example? If not, why?
119627
Let f : A → B and g : B → C be bijections. Prove that g◦f : A → C is a bijection  Can someone show me the steps I should take to solve this problem?
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...