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Oct
31
revised If some vectors in $\mathbb Q^n$ are linearly independent over $\mathbb Q$ , then are they also linearly independent over $\mathbb C$?
added 6 characters in body; edited title
Oct
30
answered Question about composition of categories
Oct
29
revised Homomorphism well defined
edited title
Oct
29
reviewed Edit How to find out the linear transformation?
Oct
29
revised How to find out the linear transformation?
added latex
Oct
29
comment Homomorphism well defined
The definition of $\theta$ as written appears to depend on whether you write $[a]_n$, or $[a+n]_n$ or $[a-42n]_n$, all of which are equal as sets - you are being asked to show that in fact you get the same answer independent of how you write the set.
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
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.