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revised Does the homology, homotopy, and geometric realization functors of a simplicial group preserve colimits?
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answered Using u-substitution in $\int \tan^3(x) \sec(x)\mathrm{d}x$
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comment If square matrix A satisfying $A^2-4A+4I=0$ does it follow that A is diagonizable?
I'm afraid it's not necessarily true that the minimal polynomial is $(\lambda - 2)^2$. For instance, if $A= 2\mathbf{I}$, where $\mathbf{I}$ is the unit matrix, then it clearly satisfies the equation, its minimal polynomial is $\lambda -2$ and it's, of course, diagonalizable. (Of course, it also satisfies the simpler equation $A - 2\mathbf{I} = 0$, but the problem doesn't preclude this possibility.)
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revised How to check F:AxI->B is continuous
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answered How to check F:AxI->B is continuous
Aug
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comment Quotient topologies and equivalence classes
Thank you, Ivo Terek.
Aug
22
comment the cone is contractible
You can get one of the best books on general topology for free here: maths.ed.ac.uk/~aar/papers/munkres2.pdf
Aug
21
comment the cone is contractible
Nope. This is not what the universal property of the quotient topology says. What it really says is the following: if $\pi \times \mathrm{id}$ is an identification and $H'$ is continuous and passes to the quotient, then, automatically the induced map $H$ is continuous. So the point here is that you need $\pi \times \mathrm{id}$ to be an identification.
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comment The homology groups of $T^2$ by Mayer-Vietoris
@mznyzgyr Because they are homotopic paths inside $A$ and $B$: you can deform continuously one into the other, just "moving" it along $A$ or $B$, can't you? Next, closed homotopic paths give homologous 1-cicles. You can find this result in any proof of the Hurewicz isomorphism theorem.
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