Session 5: Division of maps: Sections and retractions

Section 3.5 Session 5: Division of maps: Sections and retractions

Determine some witty remark to put here later.

Example 3.5.1. Exercise 1.

Show that... if \(fa_1 = fa_2\) then \(ha_1 = ha_2\text{...}\)

Solution.

For part (a), the hint suggests contradiction might be a good approach. Let’s assume that \(h = g \circ f\) and there exists some pair of points \(\mathbf{1} \longrightarrow A\) such that \(fa_1 = fa_2\) but \(ha_1 \neq ha_2\text{.}\)

Figure 3.5.2. Diagram of determination problem (a) in Session 5 Exercise 1

In this situation, there’s no possible map \(g\) here with \(h = g \circ f\) because \(g\) can’t send possibly send \(b\) to two different destinations (\(ha_1\) and \(ha_2\)).

Now let’s consider the converse relationship in (b) where we know that \((fa_1 = fa_2) \Rightarrow (ha_1 = ha_2)\) for all \(a_1,a_2 \in A\text{.}\)

Figure 3.5.3. Diagram of determination problem (b) in Session 5 Exercise 1

In this situation, we can construct a map \(g\) that has our desired property \(g \circ f = h\text{.}\) I’d imagine it like taking an image of the maps \(f\) and \(h\) over the set \(A = \set{a_1,a_2,a_3,...}\) and defining \(g\) as by the collection of arrows that have an origin at \(fa_i\) and destination \(ha_i\text{.}\)

Obviously defining \(g\) in this way will potentially leave the possibility of arrows having duplicates, but the property \((fa_1 = fa_2) \Rightarrow (ha_1 = ha_2)\) ensures that if any two arrows share the same origin they must share the same destination.

From a meta-perspective, it makes sense that the converse holds true for use to this practically as a definition. When we say "\(h\) is determined by \(f\)", we want it to be true if and only if it satisfies the properties we’ve defined it to have.

Example 3.5.4. Exercise 2.

Show that... there is at least one \(b\) in \(B\) for which \(h(a) = g(b)\text{...}\)

Solution.

I think the key idea in part (a) here is that the point \(f(a)\) precisely satisfies the properties we want for establishing the existence of \(b\text{.}\) As long as there is some map \(f\) with \(g \circ f = h\) then we can define \(b = f(a)\) and it follows directly that \(h(a) = (g \circ f)(a) = g(f(a)) = g(b)\text{.}\)

Proving the converse in (b) seems a little trickier. I think I’d want to approach this like I did with "Exercise 1" and consider making an image by applying \(g\) to each element \(b_1,b_2,... \in \mathbf{1} \longrightarrow B\text{.}\)

If we already know that for each \(gb_i\) there is at least one \(a_i\) such that \(ha_i = gb_i\text{,}\) we can begin to construct a map \(f\) by creating a collection of arrows with origin \(a_i\) and destination \(b_i\text{.}\)

Figure 3.5.5. Diagram of problem in Session 5 Exercise 2

The above definition of \(f\) given by \(f(a_i)=b_i\) should satisfy the property \(h = (g \circ f)\) since \((g \circ f)(a_i) = g(f(a_i)) = g(b_i) = h(a_i)\text{.}\) We can rest assured we’re not missing any origin points in \(A\) because we’ve already assumed that each \(a_i \in A\) each corresponds to at least one \(b_i \in B\text{,}\) and we can safely assume that no \(a_i\) has more than one destination in \(B\) because we found them by iterating over \(B\) to begin with.

Example 3.5.6. Exercise 3.

Draw... all the sections of \(f\text{.}\)

Solution.

Let’s start here by reproducing the diagram of \(f\text{:}\)

Figure 3.5.7. Internal diagram of map \(f\) in Session 5 Exercise 3

Since there’s only two points in \(B\text{,}\) we’re need to have two arrows with one originating at each. The left arrow needs to have one of four possible destinations, since there are four points in A that get mapped there by \(f\text{.}\) The right arrow needs to have one of two possible destination from the points stacked above it. Since these choices are independent, we’re looking for a total of \(4 \times 2 = 8\) maps.

Figure 3.5.8. Possible sections of \(f\) in Session 5 Exercise 3

Upon finishing the exercises, I’ve chosen to not to retract my sarcastic pun above.

Okay, yeah, that’s partly because I’m lazy, but I think it’s also a good metaphor for what I learned about this week. Writing these posts is a multi-stage process that resembles a sequence of map compositions. I read (at least some of) the section, work out the exercises, then come back and revise. What you see at the end is the composition, but not necessarily each step along the way. At the same time, some qualities of the end product itself may restrict the process leading to it but others leave flexibility.

This week’s reading seemed to be about exactly how flexible these maps could be before they start to lose information.

I could have removed the remark at the start of this and replaced it with something else. Yet, the fact that I’m now discussing it means it must be there.

This seems exceptionally well fitting to how my proofs this week worked. I’m essentially arguing that these maps exists because I can describe a set of rules to construct them. Whether you believe me or not, well, that’s another story...

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