Timeline for Useless question on rank
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 13, 2010 at 4:56 | history | edited | Tom Goodwillie | CC BY-SA 2.5 |
added 4 characters in body
|
| Jul 13, 2010 at 4:04 | history | edited | Tom Goodwillie | CC BY-SA 2.5 |
added 58 characters in body
|
| Jul 12, 2010 at 22:29 | vote | accept | ashpool | ||
| Jul 12, 2010 at 22:27 | comment | added | Tom Goodwillie | @KConrad: All I am saying is that I prefer one of two possible definitions of "constant map from $X$ to $Y$ -- definitions which only disagree when both $X$ and $Y$ are empty. (1) there exists $y$ such that for every $x$ $f(x)=y$, (2) when $f(x_1)=f(x_2)$ always. | |
| Jul 12, 2010 at 22:24 | comment | added | Tom Goodwillie | @Mariano: Your argument proves that every constant map from the empty set to the empty set is equal to the identity map. That does not contradict my statement that the identity map from the empty set to itself is not a constant map. | |
| Jul 12, 2010 at 22:22 | comment | added | Tom Goodwillie | Sorry, folks. When I edited my answer I took out this really self-indulgent second part of the answer, in which I pointed out that two constant functions from $X$ to $Y$ determined by two different elements of $Y$ are the same if $X$ is empty. I went on to assert that the identity map from the empty set to itself is not a constant function because there is no $y$ such that for every $x$ $f(x)=y$. | |
| Jul 12, 2010 at 22:13 | comment | added | KConrad | But you can't give me two different $y$'s that are both values of $f$, so in that sense how can $f$ be nonconstant? | |
| Jul 12, 2010 at 22:13 | history | edited | Tom Goodwillie | CC BY-SA 2.5 |
added 566 characters in body
|
| Jul 12, 2010 at 22:00 | comment | added | Mariano Suárez-Álvarez | Well «$\forall y\in\emptyset,\forall x\in\emptyset, f(x)=y$» is true, no? | |
| Jul 12, 2010 at 21:56 | history | answered | Tom Goodwillie | CC BY-SA 2.5 |