Finite connected poset is contractible

algebraic-topologyorder-theorysimplicial-stuff

Given $P$ a finite connected poset is this contractible?

In my intuition the answer would be yes: I cannot really picture such a poset which is not contractible.

I know that if $P$ has a maximum or minimum the claim is true, but what about the case where there are only maximal and minimal elements?

I tried to argue as follows: taken by absurd a non-trivial element in $\pi_n(|P|)$ I would like to prove that it comes from the realization of a map of simplicial sets $\Delta_n/\partial \Delta_n \rightarrow NP$ which should correspond to a chain of morphisms $p_0 \leftrightarrow p_1 \leftrightarrow \dots \leftrightarrow p_n$ and now I should use transitivity of the order to produce an extension to a map $\Delta_n \rightarrow NP$.

For example in the situation $n=2$ I should get a map taking three points $p_0, p_1, p_2$ and morphisms $p_0 < p_1$ and $p_1 < p_2$ thus I get $p_0 < p_2$ and extend to a 3-simplex $\Delta_3 \rightarrow NP$.

But I am not sure about how to formalise this and if it actually works.

Best Answer

I don't see why a finite connected poset needs to be contractible. In fact, they can capture the homotopy type of any (connected) finite simplicial complex. Consider for concreteness the poset with four points $p_1, p_2, p_3, p_4$ such that $p_1$ and $p_2$ are minimal and $p_3$ and $p_4$ are maximal. The realization of this poset is the circle $S^1$, which is not contractible.

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