Intereting Posts

sum of series using mean value theorem
Rank of free submodules of a free module over a commutative ring.
Prove that: $\sum\limits_{n=1}^\infty \dfrac{n^2}{2^n}=6$.
Number of Multiples of $10^{44}$ that divide $ 10^{55}$
Partition function- without duplicates
Find all idempotent elements in the group algebra $\mathbb CC_3$
Commutator subgroup and subgroup generated by square.
$\epsilon$-$\delta$ limit proof, $\lim_{x \to 2} \frac{x^{2}-2x+9}{x+1}$
Finding cosets of $(\mathbb{Z}_2\times \mathbb{Z}_4)/\langle (1,2)\rangle$
The sum of three colinear rational points is equal to $O$
6-letter permutations in MISSISSIPPI
A condition of abelian groups related to an automorphism
Analytic floor function, why this seems to work?
Group action and covering spaces
Normal subgroups of the symmetric group $S_N$

I’m reading up on how to prove if a function (represented by a formula) is one-to-one or onto, and I’m having some trouble understanding.

To prove if a function is one-to-one, it says that I have to show that for elements $a$ and $b$ in set $A$, if $f(a) = f(b)$, then $a = b$. I understand this to mean that if two elements in a domain map to the the same element in a codomain, then for the function to be one-to-one, they must be the same element because by definition, a one-to-one function has at most one element in the domain mapped to a particular element in the co-domain. Did I understand this correctly?

Then to prove that the function is onto, I’m reading an example that says “let’s prove that $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 5x+2$ is onto, where $\mathbb{R}$ denotes the real numbers. We let $y$ be a typical element of the codomain and set up the equation $y =f(x)$. then, $y = 5x+2$ and solving for $x$ we get $x ={y-2\over 5}$. Since $y$ is a real number, then ${y-2\over 5}$ is a real number and $f({y-2\over 5})=5({y-2\over 5})+2=y.$

- Finding the range of $f(x) = 1/((x-1)(x-2))$
- Is there a function $f\colon\mathbb{R}\to\mathbb{R}$ such that every non-empty open interval is mapped onto $\mathbb{R}$?
- Left Inverse: An Analysis on Injectivity
- Inverse function theorem question - multivariable calculus
- Compare growth rate of functions
- Why is an empty function considered a function?

I’m not really seeing how that proves anything, so can anybody explain this to me?

- Non-differentiability in $\mathbb R\setminus\mathbb Q$ of the modification of the Thomae's function
- How to show that this function is bijective
- Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$
- Degree of a function
- A question concerning measurability of a function
- Differentiable $f$ such that the set of translates of multiples of $f$ is a vector space of dimension two
- Let $f:\mathbb{R} \to \mathbb{R}$ be a function, so that $f(x)=\lfloor{2\lceil{\frac x2}\rceil}+\frac 12 \rfloor$ for every $x \in \mathbb{R}$.
- Is the function $f:\mathbb{R}^2\to\mathbb{R}^2$, where $f(x,y)=(x+y,x)$, one-to-one, onto, both?
- If the set of values , for which a function has positive derivative , is dense then is the function increasing?
- A real function which is additive but not homogenous

Yes, your understanding of a ** one-to-one** function is correct.

A function is ** onto** if and only if for every $y$ in the codomain, there is an $x$ in the domain such that $f(x) = y$.

So in the example you give, $f:\mathbb R \to \mathbb R,\quad f(x) = 5x+2$, the domain and codomain are the same set: $\mathbb R.\;$ Since, for every real number $y\in \mathbb R,\,$ there is an $\,x\in \mathbb R\,$ such that $f(x) = y$, the function is onto. The example you include shows an explicit way to determine which $x$ maps to a particular $y$, by solving for $x$ in terms of $y.$ That way, we can pick any $y$, solve for $f'(y) = x$, and know the value of $x$ which the original function maps to that $y$.

Side note:

Note that $f'(y) = f^{-1}(x)$ when we swap variables. We are guaranteed that every function $f$ that is onto and one-to-one has an inverse $f^{-1}$, a function such that $f(f^{-1}(x)) = f^{-1}(f(x)) = x$.

A function $f:A\rightarrow B$ is one-to-one if whenever $f(x)=f(y)$, where $x,y \in A$, then $x=y$. So, assume that $f(x)=f(y)$ where $x,y \in A$, and from this assumption deduce that $x=y$.

A function $f: A\rightarrow B$ is onto if every element of the codomain $B$ is the image of some element of $A$. Let $y\in B$. We can show that there exists $x\in A$ such that $f(x)=y$. Choose $x=f^{-1}(y)$ and so $f(f^{-1}(y))=y$. So for all $y\in B$, there exists an $x\in A$ such that $f(x)=y$.

You can imagine that a function of X to Y is injective when you can “enter” a copy of X to Y.

$f\colon X\to Y$ is injective if and only if:

- $x\neq y\Rightarrow f(x)\neq f(y)$, or
- If $f(x)=f(y)\Rightarrow x=y$

Intuition says that you can have a replica of $X$ in $Y$, it means for all $x\in X$ there are a $y\in Y$ which $f(x)=y$ and there are not other $x’$ with the same statement, $\therefore$ Y contains a copy of the set X

**Note**

- It works for $X\to Y$, if we want something similar from $Y\to X$ it is called surjective.
- Is possible that for some $y\in Y, \not{\exists}x\in X$ such $f(x)=y$

- Is there any intuitive understanding of normal subgroup?
- variance of number of divisors
- Proving Cantor's theorem
- Why is it not true that $\int_{-\infty}^\infty{\rm} x\,dx=0 \, $ given that x is an odd function?
- Showing that $\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1$
- Heine-Borel implies Bolzano-Weierstrass theorem
- Atiyah – Macdonald Exericse 9.7 via Localization
- The Adjacency Matrix of Symmetric Differences of any Subset of Faces has an Eigenvalue of $2$…?
- orthonormal system in a Hilbert space
- Finding the Gradient from these functions.
- One approach to showing the total variation of an absolutely continuous function f is the integral of |f'|
- How to evaluate this limit: $\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac12$?
- Proving Undecidability of first order logic without first proving it for arithmetic.
- Does local convexity imply global convexity?
- When can we not treat differentials as fractions? And when is it perfectly OK?