Section 3.11 Summary/Quiz on Opposed Maps
After reading over the How to solve the quiz problems section, I think I was mostly on track. A "fussy professor" might have taken issue with me, but since I worked through these last two sections on crowded 8-hour flights I deserve to cut myself a little slack.
The authors included one more "quiz" before the "test" and I think I’m about ready for it. It seemed like the answers to this quiz were meant as more of a study-guide than anything else.
Example 3.11.1. Problem 1.
Given two maps...
Solution.
I went with always and endomap. As long as the domain and codomain of the maps match up, there’s no reason why we shouldn’t be able to compose them. Once we do compose them, the resulting maps have the property that the codomain and domain are the same: \(A \xrightarrow{g \circ f} A\) and \(B \xrightarrow{f \circ g} B\text{.}\) This definition was repeated just a few pages prior.
Example 3.11.2. Problem 2.
If we know that \(g\text{...}\)
Solution.
I went with \(1_A\), endomap, idempotent, and \((g \circ f) \circ (g \circ f) = g \circ f\). I don’t know how many times I referred back to Article II for the definitions of retraction and section on page 49 but I think it’s finally starting to sink in. The notion that "only endomaps have a chance to be idempotent" was emphasized in the quiz solution to 2(a).
Example 3.11.3. Problem 3.
If we even know that \(f\text{...}\)
Solution.
I went with \(1_B\) and \(g\). One of the big results from Article II Exercise 2 was that if an inverse map exists then it must be unique. Stating that \(f\) is an isomorphism is equivalent to saying that the section and retraction are uniquely defined by the inverse \(f^{-1}\text{.}\)
Example 3.11.4. Problem 4.
Going back to 0...
Solution.
I chose could be different from in both cases. In the "Composition of opposed maps" on the previous page they use the maps \(g = mother\) and \(f = father\) as an example. With those definitions, \(f \circ g \circ f(me)\) would be "the father of my paternal grandmother" which is clearly a different person than "my father" \(f(me)\) so \(f \circ g \circ f \neq f\text{.}\) Likewise, the map \(f \circ g\) would be like a relation of "maternal grandfather". Clearly the "maternal grandfather of my maternal grandfather" described by \(f \circ g \circ f \circ g(me)\) would be a different person than "my maternal grandfather" \(f \circ g(me)\text{.}\) Since we only need one counter-example to prove maps different, QED.
