Timeline for Is monomorphism going in both directions sufficient for isomorphism?
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2010 at 0:34 | answer | added | Jérôme JEAN-CHARLES | timeline score: 1 | |
| Jul 18, 2010 at 21:38 | comment | added | Seamus | Right, so the problem is more that I'm misunderstanding what sort of work the categorical subobject idea is doing. But I'm glad I was right about monomorphisms... | |
| Jul 18, 2010 at 21:09 | comment | added | Mikael Vejdemo-Johansson | @seamus, the definition of subobject is slightly more subtle than what you reproduce it as; a subobject is an equivalence class of monomorphisms $S\to X$, where equivalence is given by the monomorphisms mutually factoring through each other; in other words, $S\to X$ and $T\to X$ represent the same subobject precisely if there are maps $S\to T$ and $T\to S$ such that $S\to X = S\to T\to X$ and $T\to X = T\to S\to X$. These maps have to be monomorphisms, and then you can take up Peter's argument above for the remainder of the argument. | |
| Jul 18, 2010 at 19:54 | comment | added | Peter LeFanu Lumsdaine | Often it's clear what map we mean, and so we can get away with leaving it unmentioned. But when it's not assumed, you may need to make it explicit! I think the intuition behind your weirdness may be something like the fact “if there are monomorphisms A → B → A, and their composite is the identity $1_A$, then these monos are isos”. (Proving this is a nice exercise!) In terms of subobjects: “if A is a s.o. of B, and B is a subobject of A, and these are compatible with our standard way of thinking of A as a subobject of itself, then A and B are isomorphic”. | |
| Jul 18, 2010 at 19:48 | comment | added | Peter LeFanu Lumsdaine | @Seamus: to see why it's not weird, contemplate the example of [0,1] and (0,1) above. A composite of monomorphisms [0,1] → (0,1) → [0,1] will end up embedding [0,1] as a proper subject of itself, eg as [1/4,3/4]. The categorical definition of subobject is more flexible than the set-theoretic — so it accommodates mathematical practice more naturally, eg “$\mathbb{N} \subset \mathbb{R}$” — but it shows that “$A$ is a subobject of $B$” isn't just a statement about $A$ and $B$, but also about some assumed map $A$ → $B$. [cont'd in next comment] | |
| Jul 18, 2010 at 19:33 | answer | added | James Freitag | timeline score: 2 | |
| Jul 18, 2010 at 18:27 | comment | added | Qiaochu Yuan | mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold | |
| Jul 18, 2010 at 18:16 | comment | added | Theo Johnson-Freyd | There is a standard game, that I have associated with Noah Snyder at Secret Blogging Seminar, but I've lost the link. The game goes: for each category you come across in nature, ask if both-ways monos implies that there is an iso. SET? Yes. TOP? No. | |
| Jul 18, 2010 at 17:41 | comment | added | Seamus | The comment about subobjects is this. The definition of a subobject of an object A is: an object B with a monomorphism from B to A. So in my example, A is a subobject of B and B a subobject of A, but they aren't isomorphic. This seems weird, right? | |
| Jul 18, 2010 at 17:30 | answer | added | Martin Brandenburg | timeline score: 2 | |
| Jul 18, 2010 at 15:56 | comment | added | Georges Elencwajg | Very nice, dan: my example with free groups seems complicated in contrast. My only consolation is that it might be an interesting result to know and that I am in the excellent company of Torsten (whom I hadn't read when I started answering) | |
| Jul 18, 2010 at 15:41 | answer | added | Georges Elencwajg | timeline score: 4 | |
| Jul 18, 2010 at 15:06 | comment | added | Joel David Hamkins | In the category of sets, however, it is true that when two sets inject into each other, then there is a bijection. This is the Cantor-Bernstein-Schroeder theorem. en.wikipedia.org/wiki/…. But it fails in many other naturally arising categories. MO member John Goodrick has investigated this quite a lot. | |
| Jul 18, 2010 at 15:00 | comment | added | Torsten Ekedahl | Another example is that free group of finite rank $>1$ contains free subgroups of any finite (and countable) rank. Hence, one may take two free groups of different finite ranks $>1$. | |
| Jul 18, 2010 at 14:58 | comment | added | Dan Petersen | Nothing is wrong, it is simply not true that monomorphisms in both directions guarantee isomorphism. A very concrete example is given by considering the intervals (0,1) and [0,1] in the category of topological spaces. I don't understand your comment about the standard definition of a subobject. | |
| Jul 18, 2010 at 14:52 | history | asked | Seamus | CC BY-SA 2.5 |