bio  website  math.cornell.edu/~hatcher 

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1d

comment 
Cap product à la Poincaré
Also, $T^*$ is a cellulation (a decomposition into cells), not a triangulation as stated in the question. 
1d

comment 
Cap product à la Poincaré
You are talking about a product that involves just homology and not cohomology, so this is not the cap product. Instead it is usually called the intersection product. A classical reference for the intersection product is the textbook by Seifert and Threlfall. A more recent textbook source is Bredon's "Topology and Geometry". Bredon attributes the intersection product to Lefschetz. The intersection product is a perfectly wellbehaved product, with the right hypotheses, so it's not clear why you use the words "evil", "fails", and "illdefined" in reference to it. 
Dec 9 
answered  Lens spaces and generalized Petersen graphs 
Nov 13 
awarded  Yearling 
Nov 4 
revised 
Reference for a fact (?) on homeomorphic knot complements
deleted 40 characters in body 
Nov 3 
awarded  Good Answer 
Oct 30 
awarded  Nice Answer 
Oct 30 
awarded  Nice Answer 
Oct 30 
answered  How to compute $\pi_{3}$ of $L(p,q)\# L(p',q')$? 
Oct 30 
revised 
Reference for a fact (?) on homeomorphic knot complements
deleted 47 characters in body 
Oct 30 
answered  Reference for a fact (?) on homeomorphic knot complements 
Oct 22 
comment 
A homological criterion for collapsibility?
The lemma as stated certainly seems false, with the counterexamples you described such as the house with two rooms, with $A$ a point. Surely K & S knew examples like this, so I wonder what they really had in mind when they stated this lemma. 
Sep 21 
comment 
Contractibility of space of embeddings of a disc
@IgorBelegradek: The formula seems to work equally well for embeddings $D^2\to {\mathbb R}^2$ and diffeomorphisms of ${\mathbb R}^2$. For embeddings the term $f(tx)$ is in effect restricting embeddings to smaller and smaller concentric disks, so one still gets embeddings, and then the $1/t$ factor reexpands to keep the derivative the identity at the origin. The limit exists, just as for diffeomorphisms, and is the identity. 
Sep 21 
comment 
Contractibility of space of embeddings of a disc
For question 2 the space is not contractible. There is a map from the space of topological embeddings to ${\mathbb R}^2  \{0\}$ given by $f\mapsto f(1,0)$ and this map splits (i.e., has a section) by considering only embeddings which are orientationpreserving Euclidean similarities. So the space is not simplyconnected. An extra condition analogous to the condition on the derivative in question 1 is needed, though it is perhaps not obvious what the best way to state this condition is. 
Sep 21 
comment 
Contractibility of space of embeddings of a disc
Please clarify what you mean by "proper". Usually a proper embedding is one which takes boundary to boundary, which in this case would force the embeddings to be diffeomorphisms or homeomorphisms, respectively, in the two cases. 
Sep 18 
awarded  Enlightened 
Sep 9 
comment 
Existence of $n$connected topological groups with $m$dimensional action extending that of $GL(m)$ on $\mathbb{R}^m$
A simple observation: $F$ must be nonorientable, since if it is orientable, then $G$, being pathconnected, must act by orientationpreserving homeomorphisms, contrary to the fact that the action of $GL(m)$ on ${\mathbb R}^m$ includes orientationreversing homeomorphisms. 
Sep 6 
comment 
“Economic” CWstructure for EilenbergMacLane spaces?
JUst to clarify: The usual construction of a homology decomposition, as in my algebraic topology book for example, shows that for each simplyconnected space $X$ with finitely generated homology groups there is a CW complex $Y$ and a weak homotopy equivalence $Y\to X$, where $Y$ has the minimum number of cells in each dimension consistent with the structure of $H_*(X;{\mathbb Z})$, namely, 1 cell for each infinite cyclic summand of the homology and 2 cells for each finite cyclic summand. This applies in particular for $K(\pi,n)$'s with $n>1$ and $\pi$ finitely generated. 
Sep 5 
comment 
Inverse cohomological isomorphisms
@Lennart Meier: Yes, thanks. Corrected now. 
Sep 5 
revised 
Inverse cohomological isomorphisms
edited body 