Notes:ToMond

Topology

You obviously know already that the quotient map, [ilmath]q:X\rightarrow W[/ilmath] is the "biggest" map (or makes [ilmath]W[/ilmath] the largest topology) such that any map [ilmath]\tilde{f} [/ilmath] where the following diagram commutes is also continuous:

[math] \begin{xy} \xymatrix{ X \ar[r]^q \ar[dr]_f & W \ar@{.>}[d]^{\tilde{f}}\\ & Y } \end{xy} [/math]

I have phrased this a bit weirdly but that's to make it parallel to tensors, I also use [ilmath]W[/ilmath] to mean "Whatever" as the elements of [ilmath]W[/ilmath] matter not.

Tensors

Here is how I see the tensor product (commonly denoted [ilmath]U\otimes V[/ilmath]) of two vector spaces:

[math] \begin{xy} \xymatrix{ {U\times V} \ar@2{->}[r]^{t} \ar@2{->}[dr]_{f} & {W=U\otimes V} \ar@{.>}[d]^{\tilde{f}} \\ & T } \end{xy} [/math]

Where:

• The double arrow denotes bilinear, and the single arrow denotes linear. Here [ilmath]W[/ilmath] is "whatever" and is a vector space. [ilmath]\tilde{f} [/ilmath] is a linear map
• The vector space [ilmath]W[/ilmath] is what we'd call the tensor product of [ilmath]U[/ilmath] and [ilmath]V[/ilmath]
• I don't need to tell you, but just for completeness, we consider no algebraic structure on the cartesian product [ilmath]U\times V[/ilmath] (do not think of it as the direct sum of vector spaces)

Tensor map

We can now state the tensor product as follows:

• For any bilinear map [ilmath]f:U\times V\rightarrow T[/ilmath] there is a (unique? - more on that later) linear map, [ilmath]\tilde{f}:W\rightarrow T[/ilmath] for which the diagram commutes, or [ilmath]f=\tilde{f}\circ t[/ilmath]

Parallels

These are almost exactly the same constructs! I put a question mark after unique for tensors because on the Factor (function) page I proved (without meaning to) that for the induced function [ilmath]\tilde{f} [/ilmath] to be unique the "quotient map" if you will, has to be surjective. Trivial for the topology case but not something I've proved for tensors yet (I've only just started thinking about them like this)

Question

I've been trying to formulate this property independently of topology or tensors, it's been a bit iffy so far.

I mentioned in the email about looking at the function we factor through as being a labeling scheme, which partitions its domain the least fine way it can such that there are enough labels for any other (Alectinuous/Continuous/ not sure on the word here) map can use these labels to create a commuting diagram. I came up with the following diagram:

[math] \begin{xy} \xymatrix{ & & & & & & & & {S_1} \ar@{.>}[ddlll]_{g_1} \ar@{.>}[dddd]^{g_2} \\ & & & & & & & & \\ X \ar[uurrrrrrrr]^{f_1} \ar[ddrrrrrrrr]_{f_3} \ar[rrrrr]^{f_2} & & & & & {S_2} \ar@{.>}[rrrdd]^{g_3} & & & \\ & & & & & & & & \\ & & & & & & & & {S_3} } \end{xy} [/math]

Sorry the diagram is a bit big, too me too long to text to be worth changing! Anyway here:

• [ilmath]X[/ilmath] is the domain of whatever we're dealing with and the [ilmath]S_i[/ilmath] are just "sets" - I've not used the "whatever" [ilmath]W[/ilmath] here.
• [ilmath]f_1[/ilmath] is the "quotient" map, or "tensor product" or whatever you want to call it. Note that the other [ilmath]f_i[/ilmath]s factor through it
• The diagram commutes, it must do for this "labeling" notion to work, that is that [ilmath]g_3\circ g_1=g_2[/ilmath] or:
  • [math]\forall x,y\in X\forall j>i[f_i(x)=f_i(y)\implies f_j(x)=f_j(y)][/math] ([ilmath]f_3[/ilmath] uses the least labels if you will)

Work so far

I've been looking to create two "families" (I say this to sidestep indexing them somehow) of functions, the [ilmath]f[/ilmath]s and the [ilmath]g[/ilmath]s if you will. I will call these [ilmath]\mathcal{F} [/ilmath] and [ilmath]\mathcal{H} [/ilmath] respectively (I chose this because it is easier to write, now I can't break the habit)

That last property above:

• [math]\forall x,y\in X\forall j>i[f_i(x)=f_i(y)\implies f_j(x)=f_j(y)][/math]

Is actually slightly weaker than it needs to be, we can actually say:

• [math]\forall x,y\in X\forall j\ge i[f_i(x)=f_i(y)\implies f_j(x)=f_j(y)][/math] - requiring we add the identity functions to [ilmath]\mathcal{H} [/ilmath]

So anyway, what I have so far:

• Let [ilmath]\mathcal{S} [/ilmath] be a family of sets (the [ilmath]S_i[/ilmath]
• Let [ilmath]X[/ilmath] be a set (the domain of the [ilmath]f[/ilmath]s)
• Let [ilmath]\mathcal{F}=\{f:X\rightarrow S\vert\ S\in\mathcal{S}\}[/ilmath]
• Let [ilmath]\mathcal{H}=\{g:A\rightarrow B\vert A,B\in\mathcal{S}\}[/ilmath] where:
  1. [math]\forall S\in\mathcal{S}, i_S\in\mathcal{H}[/math]
  2. [math]\forall g,h\in\mathcal{H}[/math] if [math]h\circ g[/math] is defined (they can compose) [math]h\circ g\in\mathcal{H}[/math] proof pending for diagram above
  3. [math]\forall f\in\mathcal{F}\ \forall g\in\mathcal{H}[/math] if [math]f\circ g[/math] is defined then [math]f\circ g\in\mathcal{F}[/math] proof pending for diagram above - ought to be trivial

Questions:

• Should the [math]g[/math] be unique yes - but this is based on intuition and requires the [ilmath]f[/ilmath]s be surjective
• Am I on the right lines?
• [ilmath]\mathcal{F} [/ilmath] is a VERY unnatural way to view the continuous functions - how would I show that the quotient topology is my largest [ilmath]f[/ilmath] if you will
• What is this called?