Types of Function

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Problem Set

Problem set 10 is now official.

Finish it by a week from Wednesday (i.e., by April 28); grade it during the following week.

Rejuvenation Day

This Thursday (April 22) is a rejuvenation day. It doesn’t affect this class directly, but some of you might have meetings scheduled with me on it. I’m willing to hold those meetings, but also encourage you to take the day as a break from academics — you’re welcome to reschedule meetings if you want.

Ranges of Compositions

The lingering conjecture from Friday: if h and k are functions that can be composed, then range( h ○ k ) is not necessarily equal to range( h ). Can you think of an example?

Recall that h ○ k = h( k(x) )

We sketched what the various domain, codomain, and range sets involved in this would look like, and how they’d be connected by k, h, and h ○ k. We thought about some specific examples of how small artificial functions might or might not get into the situation where range( h ○ k ) ≠ range( h ):

Eventually we realized that the key requirement is that k not map to every value in domain(h), i.e., that range(k) is a proper subset of domain(h). It’s an exercise for the listener/reader to think of specific examples of common functions that would use this idea to achieve range(h ○ k) ≠ range(h). In doing this, realize that range(k) ⊂ domain(h) is necessary for range( h ○ k ) ≠ range( h ), but not sufficient, i.e., even if you satisfy the requirement that range(k) ⊂ domain(h), you still might get range( h ○ k ) = range( h ). But you’ll never get the two ranges to be different if you don’t have range(k) ⊂ domain(h).

Injections, Surjections, Bijections

Based on section 6.3 of the textbook and the function types discussion.

For example, consider f : ℝ×ℝ → ℝ×ℝ defined by f(x,y) = (x+y, x-y). Is it an injection? a surjection? a bijection?

Examples to help understand f: f(1,1) = (2,0), f(2,4) = (6,-2)

Is f an injection?

“Injection” means “1-to-1,” i.e., that for each output there’s a unique input that produces it.

So one way to determine whether f is an injection is to suppose that f(a,b) = f(c,d).

Then if that requires (a,b) = (c,d), i.e., a = c and b = d, then f is an injection because in fact there’s only one pair (a,b) that produces f(a,b) aka f(c,d). This idea of assuming that two values produce the same output from a function and then showing that those two values must be equal is a powerful way of proving that a function is an injection in general.

On the other  hand, if (a,b) ≠ (c,d), i.e., a ≠ c or b ≠ d, then f is not an injection because there are 2 distinct inputs that produce the same output.

Next

Finish determining, whether f is a surjection and/or bijection, and proving all the results.

There’s no new reading, but review section 6.3.

Also take a look at the new question added to the injections, etc. discussion in order to start thinking about this problem.

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