Justin Young

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6h	comment	Spheres in different dimension are not homotopy equivalent Yes, but it is true if the embedding is a trivial cofibration. That, of course, uses the machinery of model categories.
23h	comment	(weak) homotopy equivalence The simplest example I know of is $X = \{0\}\cup \{1/n \vert n\ge 1\} \subset \mathbb R$ as a sub space, and $Y$ a countable discrete space, then the obvious map $Y \to X$ is a weak equivalence, but you can prove that there can be no homotopy equivalence essentially because $1/n \to 0$.
1d	comment	algebraic or homotopical proof for Kakutani fixed point theorem See: mathoverflow.net/questions/22294/… .
2d	comment	Examples of a non-finitely generated $\mathbb Z$-module The question can be reformulated to asking whether there exist an abelian group $M$ such that $\mathbb Q \to \mathbb Q/\mathbb Z$ is an $M$- colocal isomorphism, i.e., $\hom (M, \mathbb Q)\to \hom (M, \mathbb Q/\mathbb Z)$ is an isomorphism. As the answer indicates, this problem seems to be very sensitive to the underlying set theory.
May 21	comment	Is there a fibration whose base, total and fibre is a Moore space? $\mathbb Q/\mathbb Z \otimes \mathbb Q /\mathbb Z = 0$, since every element in $\mathbb Q/\mathbb Z$ is torsion and $\mathbb Q/\mathbb Z$ is divisible. Then, since $\mathbb Q$ is flat, we find that $Tor(\mathbb Q/\mathbb Z, \mathbb Q/\mathbb Z) = \mathbb Q/\mathbb Z$.
May 13	comment	Is there a fibration whose base, total and fibre is a Moore space? I don't think $G$ has to be free here. We could also have $G = \mathbb Q/\mathbb Z$, for example.
May 8	comment	$SU(n)$ is simply connected (proof without fibrations, $n>2$) There is a cell structure on $SU(n)$ with a single zero cell and no one cells: projecteuclid.org/euclid.pja/1195525543. This implies that $SU(n)$ is simply connected, but it takes a bit of work.
May 7	comment	$\pi_1(X)$ finite, show $f:X \to S^1$ is nullhomotopic Plus, this shows that for $X$ a CW-complex, one only needs to assume that $H_1(X)$ is finite, which is strictly weaker.
May 5	comment	Application of Kunneth formula to chain maps (Hatcher exercise) "precisely the same" is a bit vague or a bit too strong. I think the idea is to show there is a bijection between chain maps $I\otimes C \to C'$ and chain homotopies $C\to C'$. So, you have the bijection in one direction, now prove the other: given a chain homotopy $C\to C'$ construct a chain map $I\otimes C \to C'$.
May 3	comment	$\pi_1(X)$ finite, show $f:X \to S^1$ is nullhomotopic Ok, but no hypotheses on X were stated, and I thought this was amusing.
May 3	answered	$\pi_1(X)$ finite, show $f:X \to S^1$ is nullhomotopic
May 3	comment	Application of Kunneth formula to chain maps (Hatcher exercise) Try verifying that $h(c) = f(e\otimes c)$ is a chain homotopy between the given chain maps by computing $dh + hd$?
May 2	answered	Is there a fibration whose base, total and fibre is a Moore space?
May 2	answered	graphs and homotopy extension property
Jan 28	comment	Calculating H_0 directly from Eilenberg-Steenrod axioms Yes, I was using my own version of the ES axioms, which apparently is not the standard one. You can at least show that for spaces of the homotopy type of CW complexes, your result holds, and there may be some kind of hint of a counterexample in the MO answer.
Jan 25	comment	Calculating H_0 directly from Eilenberg-Steenrod axioms I don't know a better way than: 1. prove from the axioms that a CW-complex with $1$ vertex, $Y$, satisfies $H_0(Y)= \mathbb Z$. 2. Show that if $X$ is path connected there is a CW-complex with $1$ vertex $Y$, and a weak equivalence $Y\to X$.
Jan 6	comment	Covering map from real line to unit circle using sine and cosine I wish Munkres had put that lemma in his book, it's extremely useful.
Jan 6	comment	a problem on fixed point on different sets under continuous mapping Part (a) follows from the intermediate value theorem applied to $f(x) - x$, so quoting the Brouwer fixed point theorem is a bit much.
Dec 17	comment	Postnikov invariants & Cohomology of EM spaces I think it's slightly better to think of the polynomial algebra and the exterior algebra as both being cases of a free dg commutative algebra on an even and odd degree generator, respectively. Then, the EM spaces give you the same kinds of algebras.
Nov 6	comment	Is the boundary map induced by a good pair $(D^2, S^1)$ equal to multiplication by $2$? Yes, I should have said $C_*(D^2, S^1)$ and mentioned excision.