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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 $f({y-2\over 5})=5({y-2\over 5})+2=y.$

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

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    $\begingroup$ Onto means that the range of the function is the entire co-domain. In this case, the graph of $f(x) = 5x+2$ takes all possible $Y-$values. $\endgroup$ – Prahlad Vaidyanathan Oct 28 '13 at 17:19
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    $\begingroup$ That means that $\exists x$ (namely, $\frac{y - 2}{5}$) such that $f(x) = y$. That is what onto means! $\endgroup$ – Don Larynx Dec 10 '13 at 0:56
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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 $f(f^{-1}(x)) = f^{-1}(f(x)) = x$.

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  • $\begingroup$ but..what I'm seeing is that if x is (y-2)/5, then f(x) = y. I don't see how that says that for EVERY x, f(x) = y, which I assume is what I'm trying to prove? $\endgroup$ – FrostyStraw Oct 28 '13 at 17:39
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    $\begingroup$ $x = \frac{y-2}{5}$. Let's just pick any $y$: take $y = 2$. Then $x = \dfrac{2-2}{5} = 0$, so we see that there is an element in the domain, namely $x = 0$, that f maps to y = 2: $f(0) = 2$. What we've shown is simply that for any y whatsoever in the codomain, there exists an x in the domain such that $f(x) = y$. $\endgroup$ – amWhy Oct 28 '13 at 17:42
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    $\begingroup$ although I guess it means that no matter what that x value represents numerically, it will give y if plugged in...hmm $\endgroup$ – FrostyStraw Oct 28 '13 at 17:43
  • $\begingroup$ @amWhy: Nice write - up +1 $\endgroup$ – Amzoti Oct 29 '13 at 1:55
  • $\begingroup$ when trying to prove that a fincion is onto, would it be formal argument to say: $f$ is onto because for every $x \in \text{image} f, \ x$ is necessarily $\in \text {dom} f$? $\endgroup$ – Jneven Aug 15 '18 at 10:48
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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 $f(x)=y$.

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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 $f(x)=y$
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