# Thomas Belulovich

Providence, RI

Age: 25

 Apr 21 answered $(x_1y_2 - x_2y_1)$: area or displacement? Apr 11 awarded Supporter Nov 25 comment 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. Nov 25 revised The wedge sum is a retract of the torus?added 58 characters in body Nov 23 awarded Nice Answer Nov 23 answered Tricks to remember Fatou's lemma Nov 19 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. Nov 18 comment Writing functions without 'x' - using only composition and partial-application of other functions?This may be of interest: haskell.org/haskellwiki/Pointfree Nov 14 comment 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. Nov 14 answered The wedge sum is a retract of the torus? Nov 14 comment $h_*$ is the trivial homomorphism, the $h$ is homotopic to a pointAlso, 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. Nov 14 comment $h_*$ is the trivial homomorphism, the $h$ is homotopic to a pointThis 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. Nov 14 answered Combinatorial proofs: having a difficult time understanding how to write them out Nov 14 comment Combinatorial proofs: having a difficult time understanding how to write them outDo you mean $\sum k\binom{n}{k}$? Nov 14 answered $h_*$ is the trivial homomorphism, the $h$ is homotopic to a point Nov 9 comment Killing successive homotopy groups via fibrationsThis 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.) Nov 6 comment 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 $C_*(D^2)/C_*(S^1)$. Nov 5 comment 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. Oct 8 comment 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? Aug 29 comment Connectivity of a map and the long exact sequence of homotopy groupsWikipedia doesn't seem to claim the sequence $(*)$ extends on both sides.