Inverse Functions vs Inverse Relations

[opus]

If we have a relation, $R$, and it's inverse, $R^{-1}$ they behave such that a point on $R$, say (a,b), corresponds to the point (b,a) on $R^{-1}$ This is a reflections across the line y=x.

This relation does not mean that $R^{-1}$ is a function. For example,

Let $R$ be $y=-x^2-5$. Then $R^{-1}$ is $x=-y^2-5$ These are both parabolas reflected across the line y=x. However, since $R$ is not one-to-one, then $R^{-1}$ is not a function.

Now, consider the two relations as functions, rather than just relations.
Then we have, $f\left(x\right)=-x^2-5$
To find the inverse function of $f$, we swap the variables x and y, then solve for y.
This results in $f^{-1}\left(x\right)=\sqrt{-x-5}$

Now I have a attached a graph of these functions with respect to one another as reference to my question:

Take a look at the domains and ranges of $R$ and $R^{-1}$.
They swap, as they are reflections of each other across the line y=x.
Domain $R$: $\left(-∞,∞\right)$, Range of $R$: $\left(-∞,-5\right]$
Domain $R^{-1}$: $\left(-∞,-5\right]$, Range of $R^{-1}$: $\left(-∞,∞\right)$Now in looking at the graph of $f^{-1}\left(x\right)$, the range is restricted to the interval $\left[0,∞\right)$, which is what makes it a function by having one input to exactly one output.So my question: When we find the inverse relation of a function, there are no restrictions. The domains and ranges swap. When we find the inverse function of a function, restrictions apply which allow the inverse to be a function.

Where and how does this "restriction take place"? Does it happen because when we swap x and y then solve for y, we execute an operation that is the opposite of the main operation of $f\left(x\right)$? And in this case, the main operation for $f^{-1}\left(x\right)$ is the square root, which in its nature has restrictions, so the inverse graph itself is then restricted by the main inverse function operator? Or does the restriction take place before we even find the inverse, by restricting the domain of $f\left(x\right)$?

 

[andrewkirk]

I like your analysis. It had not occurred to me that a function is a special case of a relation. I had thought of them as completely separate beasts.

I think the statement that starts 'When we find the inverse function of a function ...' causes trouble. The 'the' needs to be 'an', because a function that is not injective will have many inverse functions. An inverse function is a function that, when composed with the original in either order, gives the identity function on the relevant domain. It follows that any function that is non-injective on a subset of its domain that has an image of infinite cardinality has infinitely many different inverse functions.

For instance, although we might think of the function $f(x)=x^2$ as having two inverse functions, giving the positive and negative branches of the square root respectively, there are actually as many inverse functions as there are subsets of the image $(0,\infty)$, because for any such subset A, there is an inverse function that gives the negative square root on A and the positive square root elsewhere. For example one inverse function gives the negative square root for all rational inputs and the positive square root for irrational inputs. Or we could swap the positive and negatives in that to get another function, or replace 'rational' by 'algebraic' or 'transcendental' or 'between 1 and 2'.

Of course if we place an additional requirement that the inverse function must be continuous, the number of possible inverse functions dramatically reduces - to two in the case of the above square function. But there is nothing in the definition of inverse function that says it must be continuous.

To address your question then, I'd say that the restriction takes place after we derive the inverse relation. A full description of the restriction might be as follows:

Every function is a relation, but not every relation is a function. Let us say that a unary function $f:D\to G$ is a manifestation of a binary relation $R\subseteq D\times G$ if the set of all first elements of the ordered pairs comprising $R$ is equal to $D$. Then if we start with a function $f:D\to G$, interpret it as a relation $R$ and take the inverse relation, naming it $R'$ (because the latex is easier to write than $R^{-1}$), every unary function that is a manifestation of the relation $R'$ is an inverse function of $f$. Let's work out how many different manifestations there are:

Let $I$ be the set of all first elements of $R'$ (second elements of $R$, note this is also the Image of $f$), which will be a subset of $G$. We know the set of all second elements of $R'$ is $D$. For $y\in G$ let $V(y)\triangleq \{x\in D\ :\ (y,x)\in R'\}$. We avoid the usual practice of calling this set $f^{-1}(y)$ to avoid making it look like an inverse function.

For each cardinality C, define

$$I_C = \{y\in I\ :\ |V(y)|=C\}$$

That is, $I_C$ is the set of all points $y\in I$ for which the pre-image of $y$ under $f$ has cardinality $C$. Let $D_C\triangleq \{x\in D\ :\ \exists y\in I_C((y,x)\in R')\}$ be the pre-image of $I_C$ under $f$.

Then $I_1=I$ iff $f$ is injective. Otherwise there will be other cardinalities represented. Let $S$ be the collection of all cardinalities $C$ such that $|I_{C}|>0$. Then the set of all inverse functions of $f$ is in bijection to the cartesian product, for $C\in S$, of sets $J_C$ of injective functions from $I_C$ to $D_C$, the pre-image of $I_C$. That is, the set of inverse functions is in bijection to:

$$U\triangleq \prod_{C\in S} \left\{g_C:I_C\to D_C\ :\ g_C\textrm{ is injective} \wedge \forall y\in I_C((y,g(y))\in R')\right\}$$

The set-theoretic representation of the inverse function that corresponds to a particular element $z$ of $U$ is the union over $C\in S$ of the $C$th components of $z$ (the $C$th component of $x$ is a bijective function from $I_C$ to $D_C$).

In short, most relations have many functions that are manifestations of them. If we convert a function into a relation, then invert the relation, it usually will not be a function, but will have many, often infinitely many, different functions that are manifestations of it. Only a few of the manifestations may be continuous.

 

[opus]

Man that's a great response, thank you! I've spent some time trying to decipher the lingo you've got there and I still don't understand any of it unfortunately. Is that something I should know in PreCalculus? I think I get the idea though. Relations have many inverses, but there is only a single inverse to it when we make it one-to-one?

 

[andrewkirk]

Is that something I should know in PreCalculus?

I'm not sure, as Pre-Calculus is an educational concept that is not used where I live. The above has nothing to do with calculus though. The key set of concepts that's needed to understand questions like this about functions and relations is set theory.

For instance, the 'cartesian product' $A\times B$ of two sets A and B is the set of all ordered pairs (a,b) such that a is in A and b is in B. Both unary functions and binary relations are just sets of ordered pairs in $A\times B$. A unary function is one that takes only one input, which is the first element of the ordered pair, and its output is the second element of the ordered pair. In the case of a function, the first set A is called the domain and the second set B is called the 'codomain' or 'range'.

A and B need not necessarily be different sets. For instance, in the function $f(x)=x^3$ the domain and codomain are usually both taken to be $\mathbb R$, ie all of the real numbers.

The 'image' of a function is the set of all outputs from it, ie the set of all second elements of the collection of ordered pairs that makes up the function. The image is a subset of the domain. If the image is all of the domain (ie the function 'covers' the codomain), the function is called 'surjective' or 'onto'.

An 'injective' function is one-to-one. A 'bijective' function is one that is both injective and surjective, so that there's a one-to-one relation between all elements of the domain and all elements of the codomain.

The 'cardinality' of a set S, indicated by |S|, is the number of elements it has. That need not be an integer, as the cardinalities also include infinity, for infinite sets, and there are actual a whole series of different 'levels' of infinity.

$\wedge$ means 'and'. $\triangleq$ means 'is defined to be equal to'. The colon : usually means 'such that'. $\in$ means 'is an element of'.

$\prod_{C\in S}\{\textrm{some set definition involving }C\}$ means the Cartesian product of all the sets defined using that definition in curly braces, for every different element C in the collection S. So if S contains only three elements $C_1,C_2,C_3$ that indicates the set of all triples whose first, second and third elements are selected from {some set definition involving $C_1$}, {some set definition involving $C_2$} and {some set definition involving $C_3$} respectively.

If there's any other terminology you're unfamiliar let me know and I'll try to explain it.

I think I get the idea though. Relations have many inverses, but there is only a single inverse to it when we make it one-to-one?

Almost. It is functions, not relations that have many inverses. A binary relation only has one inverse, which is the relation obtained by swapping the order of the elements in each pair. By the way, I have not come across the term 'inverse' used about relations rather than functions, but the way you are using it is coherent and well-defined so I'm adopting your terminology there.

I would say instead that a function may have many inverses, but if it is injective (one-to-one) it has only one inverse.

 

[Mark44]

Now, consider the two relations as functions, rather than just relations.
Then we have, $f\left(x\right)=-x^2-5$
To find the inverse function of $f$, we swap the variables x and y, then solve for y.
This results in $f^{-1}\left(x\right)=\sqrt{-x-5}$

Well, sort of.
Your function f is not one-to-one, which is obvious from the graph, so an inverse that is a function doesn't exist. To obtain an inverse function, you need to restrict the domain so that the resulting restricted-domain function is one-to-one.

In your example, a possible restriction is $f(x) = y = -x^2 - 5, x \ge 0$. Another possibility is to restrict the domain to {x | x <= 0}, in which case you get a different inverse function.

Many or most precalc textbooks show the process of finding an inverse starting with the step of switching x and y. What I've found is that many students think, mistakenly, that doing the operation is all they need to do. In reality, given y = f(x), where f is one-to-one (possibly restricting the domain to make it one-to-one), here's the process:

1. Solve for x in terms of y.
2. Stop.

For the example above.
$y = -x^2 - 5, x \ge 0$
$\Rightarrow -x^2 = y + 5$
$\Rightarrow x^2 = -y - 5$
$\Rightarrow x = +\sqrt{-y - 5} = f^{-1}(y)$
Because of our restriction on x, we need to choose the positive square root in the last step above.

At the precalc level, much emphasis is given to writing functions as functions of the independent variable, usually x. For the above, if the inverse has to be given as a function of x, it's a simple matter to write it as $y = f^{-1}(x) = \sqrt{-x - 5}$. At the level of calculus, switching variables is counterproductive, as you usually don't want a completely different graph -- you just want to be able to write an equation from the other perspective. That is, instead of some equation y = g(x), you want to have x as a function of y. IOW, you want $x = g^{-1}(y)$. This sort of thing happens all the time once you get to integration, finding the areas of regions between to curves.