A function \(f : A \to B\) is said to be injective (or one-to-one, or 1-1) if for any \(x,y \in A\text{,}\) \(f(x) = f(y)\) implies \(x = y\text{.}\) Alternatively, we can use the contrapositive formulation: \(x \not= y\) implies \(f(x) \not= f(y)\text{,}\) although in practice usually the former is more effective.
Note: injective functions are precisely those functions \(f\) whose inverse relation \(f^{-1}\) is also a function. You should prove this to yourself as an exercise.
Let \(f : A \to B\) be a function and \(f^{-1}\) its inverse relation. Then \(f\) is injective if and only if the restriction \(f^{-1}|_{\range(f)}\) is a function.
A function \(f : A \to B\) is said to be surjective (or onto) if \(\range(f) = B\text{.}\) That is, for every \(b \in B\) there is some \(a \in A\) for which \(f(a) = b\text{.}\)
A function \(f : A \to B\) is said to be bijective (or one-to-one and onto) if it is both injective and surjective. We also say that \(f\) is a one-to-one correspondence.
The composition of injective functions is injective and the compositions of surjective functions is surjective, thus the composition of bijective functions is bijective. That is, let \(f: A \to B\) and \(g: B \to C\text{.}\)
Suppose \(f,g\) are injective and suppose \((g \circ f)(x) = (g \circ f)(y)\text{.}\) That means \(g(f(x)) = g(f(y))\text{.}\) Since \(g\) is injective, \(f(x) = f(y)\text{.}\) Since \(f\) is injective, \(x = y\text{.}\) Thus \(g \circ f\) is injective.
Suppose \(f,g\) are surjective and suppose \(z \in C\text{.}\) Since \(g\) is surjective, there exists some \(y \in B\) with \(g(y) = z\text{.}\) Since \(f\) is surjective, there exists some \(x \in A\) with \(f(x) = y\text{.}\) Therefore \(z = g(f(x)) = (g \circ f)(x)\) and so \(z \in \range(g \circ f)\text{.}\) Thus \(g \circ f\) is surjective.
If \(f,g\) are bijective then \(g \circ f\) is also bijective by what we have already proven.
Determine whether or not the restriction of an injective function is injective. If it is, prove your result. If it isn't, provide a counterexample.
As we established earlier, if \(f : A \to B\) is injective, then the restriction of the inverse relation \(f^{-1}|_{\range(f)} : \range(f) \to A\) is a function. Moreover, if \(f : A \to B\) is bijective, then \(\range(f) = B\text{,}\) and so the inverse relation \(f^{-1} : B \to A\) is a function itself. The next theorem says that even more is true: if \(f: A \to B\) is bijective, then \(f^{-1} : B \to A\) is also bijective.
Suppose \(f : A \to B\) is bijective, then the inverse function \(f^{-1} : B \to A\) is also bijective.
Suppose \(b,y \in B\) with \(f^{-1}(b) = a = f^{-1}(y)\text{.}\) Thus \(b = f(a) = y\text{,}\) so \(f^{-1}\) is injective.
Now suppose \(a \in A\) and let \(b = f(a)\text{.}\) Then \(f^{-1}(b) = a\text{.}\) Thus \(A = \range(f^{-1})\) and so \(f^{-1}\) is surjective.
Let \(A\) be a nonempty set. A permutation of \(A\) is a bijection from \(A\) to itself.
Notice that we now have two different instances of the word permutation, doesn't that seem confusing? Well, let's see that they aren't that different after all. Let \(A\) be a nonempty finite set with \(n\) elements \(a_1,\ldots,a_n\text{.}\) Then let \(f : A \to A\) be a permutation (as defined above). Then \(f(a_1),\ldots,f(a_n)\) is some ordering of the elements of \(A\text{,}\) i.e. a permutation in the sense of combinatorics. Notice that nothing in this list is repeated (because \(f\) is injective) and every element of \(A\) is listed (because \(f\) is surjective). So, every function permutation gives us a combinatorial permutation.
However, we also need to go the other way. Let \(b_1,\ldots,b_n\) be a (combinatorial) permutation of the elements of \(A\text{.}\) Define a function \(f: A \to A\) by \(f(a_1) = b_1\text{.}\) Since any element of \(A\) is only listed once in the list \(b_1,\ldots,b_n\text{,}\) then \(f\) is injective. Since every element of \(A\) occurs somewhere in the list \(b_1,\ldots,b_n\text{,}\) then \(f\) is surjective.
So, what is the difference between a combinatorial permutation and a function permutation? Well, two things: one is the way we think about it, but here each viewpoint provides some perspective on the other. However, the other difference is perhaps much more interesting: combinatorial permutations can only be applied to finite sets, while function permutations can apply even to infinite sets! This means that a permutation \(f : \mathbb{N} \to \mathbb{N}\) can be thought of as “reordering” the elements of \(\mathbb{N}\text{.}\)
Let \(A\) be a nonempty set.
All of these statements follow directly from already proven results.
The above theorem is probably one of the most important we have encountered. Basically, it says that the permutations of a set \(A\) form a mathematical structure called a group. A group is just a set of things (in this case, permutations) together with a binary operation (in this case, composition of functions) that satisfy a few properties:
Chances are, you have never heard of a group, but they are a fundamental tool in modern mathematics, and they are the foundation of modern algebra. Groups will be the sole object of study for the entirety of MATH-320!
Groups were invented (or discovered, depending on your metamathematical philosophy) by Évariste Galois, a French mathematician who died in a duel (over a girl) at the age of 20 on 31 May, 1832, during the height of the French revolution. Galois invented groups in order to solve, or rather, not to solve an interesting open problem.
In high school algebra, you learn that a quadratic equation of the form \(ax^2 + bx + c = 0\) has two (or one repeated) solutions of the form \(x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\text{,}\) and these solutions always exist provided we allow for complex numbers. This formula was known even to the Greeks, although they dismissed the complex solutions.
There is a similar, albeit significanlty more complicated, fomula for the solutions of a cubic equation \(ax^3 + bx^2 + cx + d = 0\) in terms of the coefficients \(a,b,c,d\) and using only the operations of addition, subtraction, multiplication, division and extraction of roots. There is another similar formula for quartic equations, but the cubic and the quartic forumlae were not discovered until the middle of the second millenia A.D.!
Then for a few hundred more years, mathematicians search for a formula to the quintic equation satisfying these same properties. Galois invented groups in order to solve this problem. Although, instead of finding a formula, he proved that no such formula exists for the quintic, or indeed for any higher degree polynomial. When we say that no such formula exists, we mean there is no formula involving only the coefficients and the operations mentioned; there are other ways to find roots of higher degree polynomials. It should be noted that Niels Henrik Abel also proved that the quintic is unsolvable, and his solution appeared earlier than that of Galois, although Abel did not generalize his result to all higher degree polynomials. It is clear, however, that Galois did not know of Abel's solution, and the idea of a group was revolutionary.