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)