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Jun
28
comment Do we have a “short five lemma” for any two of the isomorphisms?
Yes, $A$ and $A'$ are the kernels of $B\to C$ and $B'\to C'$, and the vertical map $B\to B'$ maps the kernel of $B\to C$ isomorphically to the kernel of $B'\to C'$.
Jun
24
comment Applications of Complex Analysis.
Laplace transforms are used in engineering, they produce the transfer functions of a mechanical system, whose poles and zeros and phase have physical meaning, but I doubt techniques from complex analysis is being used regularly.
Jun
18
comment Looking For a Neat Proof of the Fact that the Grassmannian Manifold is Hausdorff
@caffeinemachine yes indeed.
Jun
12
comment Existence of diffeomorphism through convergence in Hausdorff distance
It's strange indeed, as the Hausdorff distance isn't a metric on open sets, you can have open sets that are distinct and have zero Hausdorff distance...
Jun
12
comment Is the sum of the following series a finite number or not? Explain. $ \sum_{k=1}^\infty \frac{5\sin^2k}{k!} $
Where are you stuck, and what have you tried?
Jun
12
revised Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
added 368 characters in body
Jun
11
comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
Wow, this is very neat! +1
Jun
11
comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
@JimBelk You are right, and in any case the process I propose doens't do what is expected, it's a little more complicated than I originally thought: the banana shaped regions should extend along the complete positive and negative orbit of a point. I think there might be a problem if $f$ isn't nice enough, in particular, it should work if all of $f$'s orbits are finite.
Jun
11
comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
Actually, one could forego the infinite composition altogether by just integrating $\sum_{n=1}^\infty X_n$.
Jun
11
comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
There is a slight problem with the final composite (it may not be finite on every compact set), but it can be fixed. You just need a collection of disjoint, banana-shaped open sets $B_n$ that contain $\phi(n)$ and $f(\phi(n))$ for all $n$ and no other lattice points, and have the $X_n$ have their supports in there. Actually, one could forego the infinite composition altogether.
Jun
11
revised Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
added 376 characters in body
Jun
11
revised Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
added 376 characters in body
Jun
11
answered Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
Jun
10
comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
Actually, I was probably wrong, I think I can see a construction.
Jun
10
comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
The bijection you present seems unrealizable, I would be shocked if the answer was yes in this case. I vote no.
Jun
9
comment Certain principle bundle structure on $\mathbb{R}^{n}\setminus \{0\}$
Ok, that's what I thought.
Jun
9
comment Certain principle bundle structure on $\mathbb{R}^{n}\setminus \{0\}$
How is the upper plane a non-abelian Lie group? EDIT: do you identify it with the orientation preserving affine automorphisms of the real line $\Bbb R$? Both are homeomrphic to $\Bbb R^*_+\times \Bbb R$.
Jun
8
comment Prove that if A is singular, then adj(A) is also singular
@user2097 Yes ^ ^, copy and paste at work.
Jun
8
comment Prove that if A is singular, then adj(A) is also singular
You can prove directly that $\mathrm{adj}(A)=0$ if $A$'s rank is at most $n-2$ ($n$ is the size of the square matrix $A$), and $\mathrm{rk}(\mathrm{adj}(A))=0$ if $\mathrm{rk}(A)=n-1$.
Jun
8
comment finite dimensional CW complex
@Berci no, I forgot about that, fixed it.