SUNY Geneseo Department of Mathematics
Wednesday, April 21
Math 239 03
Spring 2021
Prof. Doug Baldwin
(No.)
Monday we were looking at the function f : ℝ×ℝ → ℝ×ℝ defined by f(x,y) = (x+y, x-y), and trying to decide whether it’s an injection, a surjection, and/or a bijection.
We realized that “injection” means “one-to-one,” i.e., that each output value from an injection corresponds to exactly one input value. So one way to decide if f is an injection is to imagine that f(a,b) = f(c,d), and see if that requires (a,b) being equal to (c,d).
How would you do that?
Start by assuming that f(a,b) = f(c,d).
About the only thing you can do with this is plug a, b, c, and d into the definition of f, to get (a+b, a-b) = (c+d, c-d).
What it means for two ordered pairs to be equal is that each component of the first is equal to the corresponding component of the second. In other words, a+b = c+d and a-b = c-d
Now suppose we add these equations to each other, i.e., add the left sides and add the right sides to get a new equation: (a+b) + (a-b) = (c+d) + (c-d)
This simplifies to 2a = 2c, or a = c. So we have one of the two equalities we need to show in order to show that f is an injection!
Now that we know a = c, we can substitute a for c in earlier equations, for example a + b = c + d becomes a + b = a + d.
Now the a’s cancel out, and we’re left with b = d.
This is the last equality we need to show, so now we’ve established that f is indeed an injection!
The above records the thinking we did in class to understand why f is an injection. It’s not a formal proof, and a formal proof would have some advantages of (hopefully) being more concise while still identifying the essential ideas. So we wrote the proof formally in LaTeX; the source document and PDF file are available through Canvas.
We started by consulting the book’s definition of “surjection.” We found that the key idea in that definition is that a surjection is a function whose codomain equals its range. Then we had to figure out what this phrase means, paraphrase it in our own words. Someone suggested that the codomain is basically the set of potential values from a function while the range is the set of actual values, so the phrase from the definition means that a surjection is a function that really does produce every value that it potentially could. We then checked that the rest of the definition is consistent with this understanding, particularly the part that defines “surjection” more formally by saying that for every y in the codomain (set B in the definition) there is an x in the domain (A) such that f(x) = y. This is in fact a formal version of our paraphrase.
The definition suggests that one way to show that a function is (or isn’t) a surjection is to find (or discover that you can’t find) an expression that gives you the element(s) of the domain that produce a generic element of the codomain.
In the case of function f, this means we could let (x,y) be any pair of real numbers, and then try to find a and b such that f(a,b) = (x,y)
Once again, start by applying the definition of f to the equation f(a,b) = (x,y): a+b = x and a - b = y.
Now we basically want to solve those equations for a and b. Start by expressing a in terms of the other variables: a = x - b = y + b
Now we know that x - b = y + b. Rearrange this to isolate b (one of the values we’re solving for):
x - y = 2b
b = (x-y)/2
Now we have an expression for b, so substitute it into one of the other equations and isolate a:
a = x - (x-y)/2
= (2x - x + y)/2
= (x+y)/2
Since we found ways to calculate a and b such that f(a,b) = (x,y) for any pair of real numbers x and y, f is a surjection! (Although to be really complete, we should verify that plugging these values into f does yield (x,y), i.e., verify that all the steps we did in working from (x,y) back to (a,b) really are reversible.)
You can also find a formal version of this proof in the LaTeX source file and PDF document from today’s class.
Finally, how about applying the definition of “bijection” to f?
This turns out to be quick, because the definition says that a bijection is just a function that is both an injection and a surjection. Since we’ve just shown that f is both, we also see that f is a bijection.
Formally, our formal versions of today’s proofs states this as a corollary of the proofs about being an injection and a surjection. See the end of the LaTeX source and PDF files.
Inverses of functions.
Please read section 6.5 of the textbook.
Please also contribute to this discussion of inverse functions.