8,899 reputation
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bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 10 months
seen 2 days ago

Postgraduate student at the University of Bath, UK, studying geometry and representation theory. Currently thinking about cluster algebras and related objects.


Oct
29
comment Homomorphism well defined
That's exactly why I said that; $[a]_n$ is a set, as you point out, but $a+Pn$ is an integer. Therefore, they are not equal. However, $[a]_n=\{a+Pn:P\in\mathbb{Z}\}$ (and so $[a]_n=[a+n]_n=[a+2n]_n=\dotsb$).
Oct
29
answered Understanding why Euler's Formula applies to planar graphs
Oct
29
comment Understanding why Euler's Formula applies to planar graphs
Then that's fine - once you draw the graph on the sphere then you have one of those. I'll assemble the comments into an answer.
Oct
29
answered Homomorphism well defined
Oct
29
revised Homomorphism well defined
added 55 characters in body
Oct
29
comment Understanding why Euler's Formula applies to planar graphs
It depends what your statement of Euler's theorem says exactly - you need to know that the formula is insensitive to how the edges and faces are curved, so that you don't have to use a homeomorphism to straighten out your subdivision of the sphere before the formula applies. (You also need to know that two homeomorphic spaces have the same Euler characteristic).
Oct
29
comment Understanding why Euler's Formula applies to planar graphs
I mean just the vertices and edges - in general, if you think of a polyhedron as being built up of pieces of various dimensions (in this setting, just dimensions $0$, $1$ and $2$ for the vertices, edges and faces), the $n$-skeleton is all the stuff of dimension at most $n$.
Oct
29
comment Definition of connected graph
You need to give the definition of a walk and a chain for this question to be answerable.
Oct
29
comment Understanding why Euler's Formula applies to planar graphs
The way you phrase things, there is no question of homeomorphism - you just drew the graph on the sphere, so the result is the same sphere (now you know which points are vertices and which lie on edges, but you haven't changed any topological data). The question is whether you know that you can apply Euler's theorem without first "flattening out the faces" - you can, but it's not totally obvious that there is a homeomorphism of the sphere with a polyhedron such that the graph is sent to the $1$-skeleton of the polyhedron.
Oct
28
comment Proof of Functions!
OK, thanks - you should find that you don't need to use that $g$ is onto in part a, which is why I found it a little strange. If that is your definition (where you should probably replace $y$ by $f(x)$ so that the function actually appears in the definition!) then you have nothing to do - give that you know $f$ and $g$ are functions, each part of the problem gives you a recipe for defining $h(x)$ uniquely given any $x\in Z$ - that's the proof.
Oct
28
comment Proof of Functions!
Small query: are you sure you have stated (a) correctly? The onto assumption isn't actually necessary for $h$ to be a function. It may also help to give the definition of "function" you are using - there are several equivalent ones, and which one you use will affect how a formal argument should be phrased.
Oct
28
revised Proof of Functions!
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Oct
27
answered Induced homomorphism, first isomorphism theorem
Oct
27
revised Induced homomorphism, first isomorphism theorem
added 49 characters in body
Oct
24
comment Is 1^2^3 = $1^{2^3}$ or $(1^2)^3$
@Murplyx While what Thomas says is true, it is just easier to read a^(b^c) than to read a^b^c and remember that (a^b)^c would be more compactly written a^bc (and hope that whoever wrote a^b^c also realised this!). This is probably what Alizter was getting at - while there might be a sensible convention for interpreting a^b^c, it doesn't come up so often, so it's nice to help people out with some brackets.
Oct
24
comment Group objects in category of $\mathcal{Set}$ are groups - How to prove it?
It is also possible to phrase the set-theoretic definition of a group in terms of maps $\mu\colon G\times G\to G$ (the multiplication), $\varepsilon\colon\{e\}\to G$ (the unit map, whose image is the identity) and $i\colon G\to G$ (inversion) which are required to satisfy various identities (usually expressed as commutative diagrams). This gives an alternative (and presumably equivalent!) definition of a group object in a category.
Oct
24
comment Group objects in category of $\mathcal{Set}$ are groups - How to prove it?
As pointed out in the answer, the points of a set $G$ can be naturally identified with the maps $X\to G$ where $X$ is any one-point set.
Oct
24
comment Group objects in category of $\mathcal{Set}$ are groups - How to prove it?
Here "defines a group" means that the set $G(Z)$ is a group under the binary operation $g\times g'=\mu\circ(g,g')$. (The notation is a bit confusing here; the first $(g,g')$ is a pair of morphisms that you're multiplying to define the group structure on $G(Z)$, whereas the $(g,g')$ in $\mu\circ(g,g')$ is the pair considered as a single map $Z\to G\times G$, by $z\mapsto(g(z),g'(z))$).
Oct
24
revised Quick question: Chern classes of Sym, Wedge, Hom, and Tensor
added 33 characters in body
Oct
24
comment Nature of the range of $e^x$
@pbs Thanks, that's clearer.