Talk:Cartesian Product of Preimage with Image of Relation is Correspondence

This is [...] false. Take $R = I_S$ for some set $S$. Then $im (R) = im^{-1} (R) = S$ but certainly $R \restriction_{S \times S} \ne S \times S$.

Problem is that $T \times T \subseteq R$ is not given. --Lord_Farin 19:15, 2 July 2012 (UTC)

Darn. I can't remember what I was thinking. Common sense saves the day. Thanks Lord_Farin. Can it be quickly changed to total relation? --Jshflynn 19:25, 2 July 2012 (UTC)

More questions; consider the relation on $\{0,1,2\}$ being $R = \{(0,1),(1,2)\}$. Then $T = \{1\}$ but $R \restriction_{T \times T} = \varnothing$. Oh, and please use the _ in my name (or just abbreviate to LF) to make it less cocky (it has grown historically rather than that I am arrogant). --Lord_Farin 19:35, 2 July 2012 (UTC)

Yes, this appears correct (after interchanging $A$ and $B$, which I already did). --Lord_Farin 19:42, 2 July 2012 (UTC)

As I suspect you are writing the proof just now, I will wait until you're done and then I will move the page to Preimage Times Image of Relation is Correspondence (unless you've got a more descriptive alternative). --Lord_Farin 19:53, 2 July 2012 (UTC)

Thanks LF. By the way is "A times B" how you voice $A \times B$? --Jshflynn 23:36, 2 July 2012 (UTC)

To be completely rigorous and to conform with existing page titles, my pref would be for "Cartesian Product of Image with Preimage of Relation is Correspondence". But it's up to you. --prime mover 06:03, 3 July 2012 (UTC)

Fair enough, let's just move it again, no problem; consistency is everything. Generally I don't voice stuff in English that much, being Dutch, but it all depends on the context: it's a nice, short, grammatically correct way to voice $\times$, but I caught myself saying '$A$ Cartesian product $B$' or '$A$ product $B$' as well (rather than some long but correct phrasing with these terms). It's sort of arbitrary anyway. --Lord_Farin 06:40, 3 July 2012 (UTC)

"Times" is okay in speech, but formally it means "this number of times" in the context of "3 times 4", for example, meaning "4, repeated 3 times", technically meaningless in context of C.P. --prime mover 06:46, 3 July 2012 (UTC)

Actually, if I wanted to voice it (I rarely don't, I haven't discussed mathematics with anybody verbally for over 5 years), I'd probably say "A cross B" using the analogy of the two sets as the axes of a cartesian coordinate plane so that one set (x lines) "crosses" the other set (y lines), but this is also a non-rigorous analogy built merely to justify the voicing. --prime mover 08:09, 3 July 2012 (UTC)

That's very impressive Prime.mover. I've only been interested in maths for 2 years. So to me that seems like quite a feat of dedication. Is there some way to extend this result to relations with an arity greater than 2? It seems clear that if you restricted each factor to its "image" you would have the largest "correspondence" possible. --Jshflynn 08:58, 3 July 2012 (UTC)

Of course it could be generalised, for example saying that a relation $R$ on $\prod_{i\in I} S_i$ is $i$-full iff $\operatorname{pr}_i (R) = S_i$. Then $\prod_{i \in I} \operatorname{pr}_i(R)$ is full, i.e. $i$-full for every $i$ (maybe there is a different term for full but I don't know it; $i$-total could also be used). I have never encountered a use for such relations, however. --Lord_Farin 09:40, 3 July 2012 (UTC)