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Apr
21 |
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answered | $(x_1y_2 - x_2y_1)$: area or displacement? |
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Apr
11 |
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awarded | Supporter |
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Nov
25 |
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The wedge sum is a retract of the torus? just added that above; the map $G \to G^{ab}$ is non-injective whenever $G$ was not itself abelian. |
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Nov
25 |
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revised |
The wedge sum is a retract of the torus? added 58 characters in body |
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Nov
23 |
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awarded | Nice Answer |
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Nov
23 |
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answered | Tricks to remember Fatou's lemma |
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Nov
19 |
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comment |
How to generalize the "freeness" in free groups? If you're just thinking of Groups and Sets for the time being, the $i$-morphisms have no choice but to be morphisms of sets (after all, $X$ is a set, and $A$ could be viewed as a set). The $f$ morphisms, however, should be morphisms of groups. You can see this by trying to prove the universal property for a free group from what you have above, and look at which conditions start to become necessary. |
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Nov
18 |
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Writing functions without 'x' - using only composition and partial-application of other functions? This may be of interest: haskell.org/haskellwiki/Pointfree |
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Nov
14 |
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The wedge sum is a retract of the torus? Are you asking if there's a general fact that $X \setminus \{p\}$ is never a retract of $X$? This is true provided $p$ has a neighborhood homeomorphic to $D^n$ for some $n$, but you'd need to use homology theory -- the computations for $\pi_1$ of a wedge of spaces and for a product of spaces are very basic by comparison. |
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Nov
14 |
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answered | The wedge sum is a retract of the torus? |
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Nov
14 |
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$h_*$ is the trivial homomorphism, the $h$ is homotopic to a point Also, it's not even necessary that $[id]$ is the generator of $\pi_1(S^1)$ for the above argument to work. If $h_*$ is trivial, it's definitely the case that $h_*([id]) = 0$, and that's all we needed. |
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Nov
14 |
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$h_*$ is the trivial homomorphism, the $h$ is homotopic to a point This is part of the theorem that $\pi_1(S^1) \cong \mathbb{Z}$. It turns out the integer associated to a class of maps $[f]$ is just the "degree" of $f$, which counts how many times $f$ loops fully around $S^1$ in a signed way. It could be unclear there aren't "degree $\frac{1}2$" maps whose square is $id$, but in fact there aren't. One way to see how this works is the idea of covering spaces. |
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Nov
14 |
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answered | Combinatorial proofs: having a difficult time understanding how to write them out |
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Nov
14 |
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Combinatorial proofs: having a difficult time understanding how to write them out Do you mean $\sum k\binom{n}{k}$? |
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Nov
14 |
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answered | $h_*$ is the trivial homomorphism, the $h$ is homotopic to a point |
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Nov
9 |
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Killing successive homotopy groups via fibrations This isn't quite what you want, but for an $(n-1)$-connected space $X$ with $\pi_n(X) = G$ you can find a map $X \to K(G,n)$ that induces an isomorphism on $\pi_n$; then you can take the homotopy fiber $F$. Then $\pi_k(F) \cong \pi_k(X)$ for $k > n$, and $\pi_k(F) = 0$ for $k \le n$. (You can then iterate this, since $F$ will be $n$-connected.) |
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Nov
6 |
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Is the boundary map induced by a good pair $(D^2, S^1)$ equal to multiplication by $2$? Actually the map $C_*(D^2) \to C_*(D^2/S^1)$ isn't surjective (take a path that starts near $\partial D^2$, runs into this boundary, then exits somewhere else -- you can do this because of the identification, but that map doesn't lift to $D^2$) but $C_*(D^2/S^1)$ is chain homotopic to $C_*(D^2)/C_*(S^1)$. |
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Nov
5 |
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Is the boundary map induced by a good pair $(D^2, S^1)$ equal to multiplication by $2$? That map $\partial$ is part of a long exact sequence; the groups before and after $\partial$ are $\widetilde{H_2}(D^2)$ and $\widetilde{H_1}(D^2)$ -- both vanish. So $\partial$ has to be an isomorphism. |
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Oct
8 |
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Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ If you have a generating function $f(x)$ for a sequence $a_n$, then $xf'(x)$ gives a generating function for $na_n$. Can you produce a generating function for $a_k = k^3$ (for $k \le n$) this way? |
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Aug
29 |
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Connectivity of a map and the long exact sequence of homotopy groups Wikipedia doesn't seem to claim the sequence $(*)$ extends on both sides. |
