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9h
 comment What is the difference between a surface and the graph of a function?
The "graph of a function" is presumably the surface comprised of points $(x,y,z)$ which satisfy a relation $z=f(x,y)$. Such a surface will pass a "vertical line test," so no two points on the surface will lie on a given vertical line. However the sphere fails this vertical line test so it is not the graph of the function, though one can graph the upper or lower hemispheres individually.
9h
 answered Is a set of prime numbers open in Furstenberg's topology?
10h
 comment Applying Möbius' theorem
The equations $z=1+i,z=1-i,z=-1-i$ determine three points, not three lines. These three points are not collinear (there is no line that passes through all three), so I don't really understand your argument.
10h
 revised Applying Möbius' theorem
deleted 15 characters in body; edited title
10h
 comment 2-colorable belongs to $\mathsf P$
[...] If at any stage, the procedure asks us to color a vertex a different color than what it already is, the graph is not $2$-colorable, but otherwise if we finish coloring the graph then it is $2$-colorable. I am not familiar with computational complexity theory so I don't know how to frame that in a way that the complexity involved in each stage is transparent, unfortunately.
10h
 comment 2-colorable belongs to $\mathsf P$
My understanding: The problem is to show that deciding whether a graph is $2$-colorable or not can be done in time bounded by a polynomial in the number of vertices $n$. You're interested in how to frame the method you have in mind, but your description of your reasoning ("the other is B and so on") seems a bit vague. Here's the method that comes to my mind: color one vertex say white (arbitrarily), then color its adjacent vertices black, then color their adjacent vertices white, and so on. [...]
15h
 answered Determining the group homomorphism in semidirect product
15h
 comment Determining the group homomorphism in semidirect product
The action of $h\in H$ on $N$ inside $N\rtimes H$ is conjugation. That is, "$h(n)=hnh^{-1}$."
15h
 comment What is the exact definition of polynomial functions?
Tangentially, an oft-overlooked point is that the evaluation map is not a ring homomorphism if $R$ is noncommutative.
16h
 revised Strong characterization of $\mathbb C$ with respect to $\mathbb R$
edited tags
18h
 answered Normal subgroup if conjugate subgroup is subset
18h
 comment Normal subgroup if conjugate subgroup is subset
It's possible Qiaochu was thinking of the something like the semidirect product of additive groups ${\bf Q}\rtimes{\bf Z}$, where $1\in\bf Z$ acts on $x\in{\bf Q}$ as multiplication by $2$. I'm pretty sure I've seen this example (with ${\bf Z}[1/2]$ instead of $\bf Q$ for some reason IIRC) in an MSE answer.
19h
 comment Group of mapping is a subset of symmetric group
@PaulS With your functions, $i(1)=2=i(2)$ so $g(i(1))=g(i(2))$. Can you extrapolate from this example to see that $i$ not injective implies $gi$ not injective?
19h
 revised Group of mapping is a subset of symmetric group
rolled back to a previous revision
20h
 revised Group of mapping is a subset of symmetric group
edited body
20h
 answered Group of mapping is a subset of symmetric group
20h
 comment Group of mapping is a subset of symmetric group
Interesting, thanks - I did not realize there could be subgroups in the monoid of functions on a set that did not contain the identity mapping.
20h
 comment Group of mapping is a subset of symmetric group
If $G$ is a group comprised of mappings then those mappings must be invertible (since every element of a group must have an inverse) hence $G$ is a subset of the collection of invertible mappings, in symbols that is $G\subseteq{\rm Sym}(X)$. The fact that all of $G$'s elements are bijective functions is automatically true, there is no "if" about it, which makes the phrasing here seem strange.
20h
 comment General solution of differential equation of order 3
Almost, $dX=(\int_0^ue(s)ds)'=e(u)$ (no $e(0)$ involved)
21h
 comment General solution of differential equation of order 3
@Vrouvrou Please read en.wikipedia.org/wiki/Integration_by_parts (I use $X$ and $Y$ where Wikipedia uses $u$ and $v$, since I already have the letters $u$ and $v$ in use.)
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