วีดีโอ

"If someobody says infinity topos or something like that"

Yes, thanks. It should have all set-indexed products. Of course I knew this, and I'm not quite sure how and why the word finite came out in the lecture. It was Friday 13th!

This is correct. Note that your answer to (2) uses the axiom of choice.

@@alexsimpson8841 Yes! That's why I used the word "choose".

Thank you for your comment. Every relation of the form you describe in your comment is indeed a monomorphism in the category Rel. However, there exist other monomorphisms in Rel that are not of this form. So the property you give does not characterise the monomorphisms.

I think it's a small fix though; I'd say it's the relations between Y and Z where every y in Y has _some_ z in Z that is related to y and only to y; i.e. different elements of Y may have overlapping images but each such image needs to have at least one "own" element. Proof sketch (for the "mono => ..." direction): by looking at the relations between 1 and Y we see that a mono needs to map every subset of Y to a different subset of Z; in particular, the images of {y}, Y\{y}, and Y all have to be different. That implies that the image of {y} cannot be a subset of the image of Y\{y}, i.e. it has to have some element not contained in the latter (which is in turn the union of the images of all the other elements).

Hehe. I was trying to understand why you started your reasoning with Y and Z. I got it. Hehehe. I prefer to use X→Y for the candidate of mono and for the domain of "test functions" I will use U. So, R: X→Y relation, using function notation, R(x)=subset of Y, For any S,T relations U→X, we must have RS=RT implies S=T Oh I will have to apply relations in subsets, so for a subset p (for points) in U, S(p)=union of S(u),u in p Then, composition of relations is (RS)(p)=R(S(p)) Yeah, I guess this works. Now, if R is mono, in particular we can apply it for relations S,T: U→X with U={u}, S=u×p, T=u×q, for any p,q subsets of X, the meaning of u×p is just u×p={(u,x): x in p} So RS(u)=RT(u) means R(p)=R(q) so if p,q have the same image by R, they have to be the same sets ... This implies, for p=domain of R and q=X that R is total, because we always have R(domain of R)=R(X) Now, for each x,y in X, x≠y, indeed we must have R(x)≠R(y) But it is more complicated than that, because for x,y,z in X, all different, we also must have R(x),R(y),R(z),R({x,y}),R({y,z}),R({z,x}),R({x,y,z}) all different! We would need to have not just, let's say k_{x,y} in Y, for each x,y in X, such that k_{x,y} is in R(x) (or R(y)) but not in R(y) (or R(x)), (which can be described by k_{x,y} in R(x) symdiff R(y), the symmetric difference, which is the union of the differences OR the union minus the intersection) but we need such k_{x,p} in Y, for any x in X and p subset of X not containing x! And for q instead of p! This can easily be described "R, as a function from Subs(X)→Subs(Y), is injective" (the mono statement "R(p)=R(q) implies p=q" is exactly that) but I want to see what this means for elements ... It is easy to understand if R is just an injective funtion. In this case, the property is automatically satisfied, because each set p={x,y,z,...} corresponds to R(p)={R(x),R(y),R(z),...} so they behave exactly the same, there is no risk to have R(p)=R(q) for p≠q. If I extend R to S = R union {(x,R(y)} with x≠y, it is NOT a mono anymore, because S({x,y}) = {R(x),R(y)} = S(x) ... But I could extend R by including (x,k), k NOT in the image of R. And when I do that, I could also include (x,k) ... Right! Because to differentiate between sets with x or y I can use R(x) and R(y)! I am starting to think this is it: R is an extension of an injective function F by points outside F's codomain. Actually, more precisely, a coextension, because I am including images. So, take F: X→Y injective for any R coextension in Y\F(X) we have R: X→Y mono in Rel Oh I think I know how to produce such an F! Starting with R, for any x in X, for X≠∅, let's define the set of x-exclusives Ex(x) := R(x)\R(X\x) we must have Ex(x)≠∅ because if Ex(x)=∅, then R(x) would be contained in R(X\x) implying R(X\x) = R(X\x) union R(x) = R(X\x union x) = R(X) R mono implies X\x=X ==> X={x} which contradicts Ex(x)=∅, because Ex(x) = R(x)\R(X\x) = R(X)\R(∅) = R(X) ≠ ∅ Now, for x≠y, the exclusives sets are clearly disjoint, by the simple reason y is in X\x, so R(y) is in R(X\x), so R(y) is NOT in Ex(x) Now, just choose one F(x) in each Ex(x) and we are done ... ✓ This also means we have Prod |Ex(x)| options to choose F by this process. Let's take an example ... X = {1,2,3} Y = {1,2,3,4,5,6,7,8,9} Let's take R defined by R(1) = {1,2,3,7} R(2) = {2,4,8} R(3) = {3,4,5,6,9} We have 7 non-empty subsets to check if R is injective on subsets of X, the first 3 are in the definition, the other are R({1,2}) = {1,2,3,4,7,8} R({2,3}) = {2,3,4,5,6,8,9} R({3,1}) = {1,2,3,4,5,6,7,9} R(X) = Y Ok, they are all different. The exclusives sets are Ex(1) = {1,7} Ex(2) = {8} Ex(3) = {5,6,9} So we can consider F: X→Y defined by F(1) = 1, F(2) = 8, F(3) = 5, and R is an extension of F. We can choose other 5 options instead of F using the exclusive sets. But they are not all options, because we could consider the injective functions in X F(x) = x² F(x) = 2x F(x) = x F(x) = x+2 F(x) = x+6 They all are corestrictions of R. There is a way to measure how much a morphism is or is not a monomorphism, right??? Anyway ... this was reallly fun! Thanks for inspiring me.

It is good to observe that there is a little detail about my previous comment that could be better: given an injective function F, if R is coextension of F by points outside the image, R is mono. But we can think more generally allowing R to be coextensions preserving the exclusive sets Ex(x), I guess. Oh I get it, for some reason I was thinking union of Ex(x) = R(X) probably because my mind was thinking R(x) instead of Ex(x) ... And now I understand why my mind was thinking about the partition R(X) = E union E⁰ Hahahaha. Yeah, R is coextension of injective functions F: X→E by points on E⁰. That's the general case. Let me invoke the example again, X = {1,2,3} Y = {1,2,3,4,5,6,7,8,9} We had R as union of injective functions, R(x) = {x,2x,x,x+2,x+6,x²} or something like that, I didn't say, but I thought R like that, or included some different element for 1 or 3, so, let me see, R(1) = {1,2,3,7} R(2) = {2,4,8} R(3) = {3,4,5,6,9} (yeah I put 4 in R(3), I would but x+1, but I forgot, I guess? Or changed my mind? I don't remember, I should be sleeping! Hahaha, it is 3am, almost 4am) I have Ex(1) = {1,7} Ex(2) = {8} Ex(3) = {5,6,9} (btw we can quickly write Ex(x) looking each element of R(x) and looking for it in the other sets, if we find it, then it is not part of Ex(x),we go to the next element, if we don't find it, it is part of Ex(x)) So if E is their union, E = {1,5,6,7,8,9} so E⁰ = R(X)\E = {2,3,4} and indeed each R(x) is extension of Ex(x) by E⁰ ... My mind started thinking that, then the part that manages sleep distracted it from the idea. Haha. There some other little observations, for example, to check each function used in the union, how they related to each other. For some reason I am thinking about stalks ... Very cool. I am very happy. Time to sleep.

No. You just shifted your frame of "interest". The task you set out, is recursive and endless.