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Mar 11, 2010 at 20:23 comment added rgrig Joel, thanks. I will say something along the lines "now we need to find a one-to-one correspondence between a subset of X and a subset of Y", where X and Y are sets described previously. I think this should be clear.
Mar 11, 2010 at 20:21 vote accept rgrig
Mar 11, 2010 at 19:46 comment added Joel David Hamkins Rgrig, but if you say "bijection of its domain with its range" or "one-to-one correspondence of its domain with its range", as I suggested, then it IS what you mean to say.
Mar 11, 2010 at 19:27 comment added rgrig @Joel: I agree. I never said you were wrong. I am just pointing out that neither 'bijection' nor 'one-to-one correspondence' mean the same as (left-unique and right-unique), so they aren't what I need to say. (I did up-vote your answer :), just so you know.)
Mar 11, 2010 at 18:55 comment added Joel David Hamkins In particular, what I mean to say is that I stand by my answer.
Mar 11, 2010 at 18:45 comment added Joel David Hamkins Well, I never said R should be a function from X to Y, or a bijection between X and Y, but rather, that it is a function on its domain, or a bijection of its domain with its range. The domain of a relation R is the set of x for which there is y with xRy, and the range is the corresponding set of y. These may not be X and Y, respectively, and this should resolve the confusion. It is completely correct to say that a relation is functional if and only if it is a function from its domain to its range, and this is why the word functional is used.
Mar 11, 2010 at 17:38 comment added rgrig Joel, I think most people take R is a function from X to Y to mean that for each $x\in X$ there is exactly one $y\in Y$ such that $xRy$; similarly, I think most people take R is a bijection between X and Y to mean that R and its inverse are both functions. Also, I think most people use one-to-one correspondence as a synonym for bijection. That is not what I want to say. What I want to say is that for each $x\in X$ there is at most one $y\in Y$ such that $xRy$ and vice-versa. (This is what the left-unique and right-unique definitions that I pointed to say.)
Mar 11, 2010 at 13:18 history answered Joel David Hamkins CC BY-SA 2.5