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- Path: sparky!uunet!snorkelwacker.mit.edu!galois!riesz!jbaez
- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: What is a manifold?
- Message-ID: <1992Nov5.221538.4456@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT Department of Mathematics, Cambridge, MA
- References: <abian.720910802@pv343f.vincent.iastate.edu> <1992Nov5.022710.10234@CSD-NewsHost.Stanford.EDU>
- Date: Thu, 5 Nov 92 22:15:38 GMT
- Lines: 34
-
- In article <1992Nov5.022710.10234@CSD-NewsHost.Stanford.EDU> pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes:
- >In article <abian.720910802@pv343f.vincent.iastate.edu> abian@iastate.edu (Alexander Abian) writes:
- >>
- >> So, MY DEFINITION is:
- >>
- >> An n-dimensional differentiable manifold M is a connected
- >> subdomain of an n+k dimensional Euclidean space such that
- >> at every point, M has an n-dimensional tangent hyperplane.
- >
- >If you require manifolds to be connected, what does coproduct in the
- >category of manifolds become? (I.e. how do you add two manifolds
- >together?) Does coproduct even exist then? Seems unlikely.
- >
- >I'm guessing connectedness doesn't damage coequalizers. Anyone know
- >for sure? John?
-
- Geez, some people wouldn't understand how to tie their shoes until it
- was explained to them in the language of categories!! :-) Let us pause
- to recall the words of Goethe: "Computer scientists are like Frenchmen:
- whatever you say to them they translate into their own language and
- forthwith it is something entirely different." (Okay, Goethe said
- "mathematicians," not computer scientists.)
-
- As far as I can tell, manifolds don't have equalizers or coequalizers.
- I have never messed with coequalizers but I guess you just turn all the
- arrows around in the definition of equalizers... so given two smooth
- maps f,g from X to Y, the coequalizer would be a manifold Z with a smooth
- map from Y to Z such that any map h from Y such that hf = hg factors uniquely
- through Z. So it seems like we are taking a quotient of Y, identifying
- the point f(x) with the point g(x) for all x in X. This quotient won't
- usually be a manifold.
-
- Of course, there are spaces that are a little nastier than manifolds but
- still interesting.
-