Let $K$ be a simplicial complex and $A$ be a subcomplex of $K$ such that $A$ is contractible. Suppose that for any simplex $\sigma\in K\setminus A$ and any vertex $v\in A$, $\sigma\cup \{v\}$ is a simplex in $K$. Is $K$ contractible? Is $K$ some kind of generalized cone? Any suggestion on reference or answer is highly appreciated. Thank you!
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2$\begingroup$ If I am not mistaken, this is called a join. See join of topological spaces and join of simplicial sets. $\endgroup$– Z. MCommented May 9, 2023 at 11:50
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$\begingroup$ Yes. I didn't recognize that! thanks a lot. $\endgroup$– Power of TopologyCommented May 9, 2023 at 12:44
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2$\begingroup$ 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.) $\endgroup$– Martin TancerCommented May 9, 2023 at 13:12
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$\begingroup$ 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 $S^1$. $\endgroup$– Power of TopologyCommented May 9, 2023 at 21:13
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1$\begingroup$ 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. $\endgroup$– Martin TancerCommented May 10, 2023 at 7:01
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