SUNY Geneseo Department of Mathematics
Monday, April 16
Math 239 01
Spring 2018
Prof. Doug Baldwin
Problem set on sets and functions: see handout for details.
“Status of Women and Under-Represented Groups in various fields of STEM”
Led by Provost Stacey Robertson
Thursday, April 19, 5:00 pm
Newton 204
Section 6.4.
Based on Progress Check 6.18. Pick one of the functions given and see how many different ways you can express it as a composition of 2 functions. How about as compositions of 3 or more functions?
For example, g(x) = cos( (2x-3) / (x2+1) ) could be written as a( b(x), c(x) ) where a(x,y) = cos( x/y ), b(x) = 2x-3, and c(x) = x2+1. This makes a neat example of composition being used with a combination of 1- and 2-variable functions.
As another example, f(x) = |x2-3| could be written as a(b(x)) where a(x) = |x| and b(x) = x2-3. Or a bit more interesting, c(d(x)) where c(x) = |x-3| and d(x) = x2.
Suppose f : A → B, g : B → C.
Is it possible for range(g ○ f) to be a proper subset of range(g) (i.e., to have range(g ○ f) ⊂ range(g))?
Yes. Consider this example: f : ℝ →ℝ defined by f(x) = x2, g : ℝ →ℝ defined by g(x) = x3. Then range(g) = ℝ, but range( g ○ f ) = ℝ* where ℝ* = [0,∞), i.e., the non-negative reals.
If it’s possible to have range(g ○ f) ⊂ range(g), what does having it happen tell you about f?
That codom(f) ≠ range(f), i.e., f is not a surjection.
State any implication deduced above as a conjecture/theorem and prove it.
Theorem: Let A, B and C be sets, and f : A → B, g : B → C be functions. Then if range(g ○ f) ⊂ range(g) then f is not a surjection.
Proof: We use the contrapositive to prove that if range(g ○ f) ⊂ range(g) then f is not a surjection. So assume that f is a surjection and use element chasing to show that range(g ○ f) = range(g). In particular, suppose x is in range(g ○ f) , i.e., x = g( f(y) ) for some y in A. But what’s important about this is that x is the result of g applied to anything, because that shows that x is in range( g ). We thus have that range(g ○ f) ⊆ range(g). In the other direction, suppose x is in range(g). Then x = g(y) for some y in B. Further, since f is a surjection, y = f(z) for some z in A. Thus x = g(f(z)), i.e., x is in range(g ○ f). So we have that range(g) ⊆ range(g ○ f). We have now shown both that range(g ○ f) ⊆ range(g) and that range(g) ⊆ range(g ○ f), so we have proven that if f is a surjection then range(g ○ f) = range(g). This also establishes the contrapositive, namely that if range(g ○ f) ≠ range(g) then f is not a surjection. Finally, note that range(g ○ f) must be a proper subset of range(g) or equal to it, so if range(g ○ f) isn’t equal to range(g), it must be a proper subset. QED.
Concrete examples of compositions and the notation for them.
Prove things about compositions by element-chasing through the appropriate function applications.
Inverse functions.
Read section 6.5.