# Everest

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 Feb22 revised Free modules are projective. edited title Feb22 revised Galois group of $x^8+2$ over $\Bbb{Q}$ deleted 8 characters in body Feb22 revised Finding the Galois group over $\Bbb{Q}$. edited title Feb21 awarded Disciplined Feb17 awarded Popular Question Feb13 comment Showing that two spaces are homotopy equivalent But is it not possible to keep making $X$ thinner and thinner until there is only one point that connects the two circles? After all, can we not make it as thin as we want? We are not cutting anything while making it thinner...so why wouldn't it be a homeomorphism? Feb13 comment Showing that two spaces are homotopy equivalent I think I misunderstood the question. I thought $S^1 \vee S^1$ was the disjoint union of $S^1$ and $S^1$, but its actually two $S^1$'s that share a point, right? However, we see that $(S^1 \times S^1) - x_0$ is homeomorphic to $S^1 \vee S^1$...because if we enlarge the hole in the square and identify the sides, we will be getting exactly $S^1 \vee S^1$. So $X$ and $Y$ are homeomorphic, which automatically implies that they are homotopy equivalent. Is that correct? Feb11 revised Showing that two spaces are homotopy equivalent deleted 81 characters in body Feb11 accepted Showing that two spaces are homotopy equivalent Feb11 comment Showing that two spaces are homotopy equivalent Oh, maybe I got what you're saying...so are you saying that it is easier to consider it as a square with opposite sides identified (and a hole inside) rather than thinking of it as a torus with a hole (even though they are basically the same thing of course)...is that what you wanted to say? Feb11 revised Showing that two spaces are homotopy equivalent deleted 425 characters in body Feb11 revised Showing that two spaces are homotopy equivalent edited body Feb11 comment Showing that two spaces are homotopy equivalent Thanks, but I googled it...and it is the same as my picture, except that it is round without edges. I'm not sure if my picture is not looking right, but if you notice in my picture, the "inside" (which is pink) is different from the "outside" (which is yellow). I colored my picture so that its clearer. Feb11 revised Showing that two spaces are homotopy equivalent added 488 characters in body Feb11 comment Showing that two spaces are homotopy equivalent Thanks a lot. Actually, that is what I was doing...but I couldn't figure it out. I wasn't thinking of it as a square...I was thinking of it as a circle (which doesn't really matter, right?). But even when I tried thinking of it as a square, it didn't make any difference. I will edit my question and post a picture to clarify what I'm thinking, ok? Feb11 asked Showing that two spaces are homotopy equivalent Feb7 accepted Showing that simple modules over finitely generated commutative algebras are 1-dimensional and isomorphic to… Feb5 comment Triangular Matrices and Simple Modules @rschwied Thanks, but I just have one more question...a free module is a module with a basis, right? But there is only one generating element (since its cyclic), and the only way we can get zero is by multiplying the generating element by the zero matrix in $R$. I'm not sure if I'm doing something wrong, but this module is finitely generated by one element, and this element is linearly independent. Isn't that the definition of basis? Feb5 accepted Triangular Matrices and Simple Modules Feb5 comment Triangular Matrices and Simple Modules I have one more question (if you don't mind). Just to be sure...are you saying my Jordan Hölder decomposition is right or wrong? I'm just asking because you didn't explicitely say that. Anyways, thanks for answering.