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Timeline for Generalized Cones

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May 10, 2023 at 12:44 comment added Power of Topology My bad. In my mind, $K\A$ is the induced subcomplex from $K$ with the vertex set $K^{(0)}\setminus A^{(0)}$. Thanks a lot!
May 10, 2023 at 7:01 comment added Martin Tancer I think that this example does not satisfy your definition. Let $a$, $b$, $c$, $d$ be the vertices of the square (with $a$ and $c$ oppposite) and $e$, $f$ be the two remaining vertices. Then $ae$ is a simplex in $K \setminus A$ and $c$ belongs to $A$. Thus, according to your condition, $ace$ should belong to $K$ as well but it does not. You probably indeed mean join of two complexes (in this case the square and two isolated points) but the definition is slightly different. With iterated cone, I simply meant taking a cone several times.
May 9, 2023 at 21:13 comment added Power of Topology What is an iterated cone pls? I think $K$ may not be contractible if $A$ is not. For example $K$ is the surface of square bipyramid and consider $A$ to be square; this is a suspension of a square, i.e. $\simeq S^2$, also a join of $S^0$ and $S^1$.
May 9, 2023 at 13:12 comment added Martin Tancer If I understand the definition, it is indeed a join but somewhat trivial one. If $\sigma$ is a simplex in $K \setminus A$, then adding $v$ again gives a simplex in $K \setminus A$. By iterating this, full simplex on $A$ belongs to the construction. Overall the result does not seem to depend on the structure (or contractibility) of $A$ but it only gives a join of the subcomplex on vertices of $K \setminus A$ with the full simplex on $A$. (I.e. an iterated cone.)
May 9, 2023 at 12:44 comment added Power of Topology Yes. I didn't recognize that! thanks a lot.
May 9, 2023 at 11:50 comment added Z. M If I am not mistaken, this is called a join. See join of topological spaces and join of simplicial sets.
S May 9, 2023 at 1:37 review First questions
May 9, 2023 at 3:58
S May 9, 2023 at 1:37 history asked Power of Topology CC BY-SA 4.0