Let $X$ be a non-contractible, $(d-1)$-connected, $d$-dimensional simplicial complex. By the theorems of Hurewicz and Whitehead, $X$ is homotopy equivalent to a wedge of $d$-spheres. Does there exist a $d$-simplex that can be removed from $X$ without decreasing the connectedness?
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$\begingroup$ If $X$ is the wedge of a 2-simplex and the boundary of a 3-simplex, remove the interior of the 2-simplex and it's not simply connected any more. $\endgroup$– John PalmieriCommented Feb 15, 2017 at 16:35
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$\begingroup$ Sorry, the question was poorly stated, I have now edited it. The question is whether there exists a $d$-simplex with the property, not whether it holds for all $d$-simplices (which is clearly false, as your example shows). $\endgroup$– user1272680Commented Feb 15, 2017 at 16:42
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2 Answers
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For a counterexample, we glue two-dimensional Moore spaces for the groups $\mathbb Z/2$ and $\mathbb Z/3$ along a common $S^1$. As a CW complex, $X$ can be realised by gluing two 2-disks into a closed loop $\gamma$, such the boundaries of the disks wind around $\gamma$ two and three times, respectively.
To get a simplicial complex, start with a triangle, then glue in a simplicial 6-gon and a simplicial 9-gon as above (both need a sufficiently fine simplicial structure for this to work). Removing a 2-simplex produces a space that is homotopy equivalent the CW complex above with one of the two disks removed, that is, to one of the two Moore spaces.
The fundamental group has a generator $[\gamma]$ and two relations $2[\gamma]=0$ and $3[\gamma]=0$, so $X$ is simply connected. If you take away one of the two disks, you loose one relation, so you loose simple connectivity.
The associated cellular chain complex looks like $$\mathbb Z\stackrel{0}\longleftarrow\mathbb Z\stackrel{(3,2)}\longleftarrow\mathbb Z^2\;,$$ and one checks that $H_2(X)\cong\mathbb Z$, so $X$ is not contractible.
answered Feb 15, 2017 at 20:16
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1$\begingroup$ This is easy to realize as an ordinary simplicial complex. Take a simplicial circle, and two simplicial discs: one wrapped twice round the circle, one wrapped three times round the circle. As you observe, this is simply connected but non-contractible. Now the removal of any 2-simplex turns the corresponding disc into a homotopy annulus, which again makes the result non-simply connected. $\endgroup$– HJRWCommented Feb 15, 2017 at 20:50
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$\begingroup$ Oh, it looks like you already did this in the edit! $\endgroup$– HJRWCommented Feb 15, 2017 at 20:53
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1$\begingroup$ I was just rewriting the answer when your comment showed up. Thanks. $\endgroup$ Commented Feb 15, 2017 at 20:53
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$\begingroup$ Many thanks for the counterexample, this is really helpful! $\endgroup$ Commented Feb 15, 2017 at 21:20
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This is just another point of view on the already-accepted answer.
Especially when working homotopy-theoretically, it's hard to think about `removing a simplex', but we can think instead about adding a simplex (or cell). This leads to the reformulation: if $S^{n-1} \to X \to \bigvee S^n$ is a cofiber sequence (so $X$ is the result of subtracting the cell from the wedge), then what can you say about the connectivity of $X$?
For $n$ sufficiently large (probably $\geq 3$ or so) this is a stable question, so we can back up the cofiber sequence to get $\bigvee S^{n-1} \to S^{n-1} \to X\to \bigvee S^n$, and we simply need the map $\bigvee S^{n-1} \to S^{n-1}$ to be surjective on integer homology. This includes, in particular, the accepted example.
answered Feb 16, 2017 at 14:28
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$\begingroup$ Indeed, I found the example by looking at the corresponding part of the homology exact sequence. But then dimension 2 was sufficient to realise it, so I had to discuss the fundamental group, too. But I am confused about the the last statement: in order to loose connectivity, we need the map to be not surjective on homology. Is that what you meant? In the example above, the map induces multiplication by 2 or 3 in $\mathbb Z$. $\endgroup$ Commented Feb 16, 2017 at 18:14