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 Apr 26 awarded Nice Answer Apr 22 awarded Nice Answer Mar 26 awarded Popular Question Feb 2 revised Does the homology, homotopy, and geometric realization functors of a simplicial group preserve colimits? added 139 characters in body Oct 30 answered Using u-substitution in $\int \tan^3(x) \sec(x)\mathrm{d}x$ Sep 15 reviewed Leave Open A simple question about ring theory Sep 15 reviewed Leave Open Is there any different between $\mathbb{C}[x]$ and $\mathbb{C}[[x]]$ Sep 5 awarded Informed Aug 30 awarded matrices Aug 27 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.) Aug 25 revised How to check F:AxI->B is continuous added 321 characters in body Aug 25 answered How to check F:AxI->B is continuous Aug 23 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. Aug 15 awarded algebraic-topology Aug 7 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. Aug 4 awarded Yearling Jul 22 awarded Nice Answer Jun 21 awarded Good Answer