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$\textbf{Definitions}$

A simplicial complex is a set $V$ called its set of vertices together with a subset $\Sigma_V \subset 2^X$ so that the sets in $\Sigma_V$ cover $X$ and is closed under taking arbitrary subsets i.e if $A \subset B \in \Sigma_V$ then $A \in \Sigma_V$.

A simplicial complex is also a topological space with set of points $V$ and topology $\Sigma$.

For two simplicial complexes with vertex sets $V$ and $W$ call a function of sets $f: V \rightarrow W$ a continuous map if it is continuous w.r.t the topologies $\Sigma_V$ and $\Sigma_W$ and a simplicial map if the image of a set in $\Sigma_V$ is always in $\Sigma_W$.

You can also associate a topological space $\mid V\mid$ to a simplicial complex called the geometric realization of $V$.

$\textbf{Questions}$

Are there any examples of simplicial maps that aren't continuous maps? What about continuous maps that aren't simplicial? Does it matter if we only restrict to finite vertex sets? This question was inspired by

Are the homotopy groups of the space $(V,\Sigma_V)$ equal to the homotopy groups of $\mid V \mid$? Are they weakly equivalent? Homotopy equivalent?

This particular question was inspired by the recent post How many homotopy types do you get from three points?

I can't 100% understand the answer because he talks about simplicial complexes being homeomorphic to the nerve of some poset, which is a simplicial set?

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    $\begingroup$ What do you mean by "A simplicial complex is also a topological space with set of points $V$ and topology $\Sigma$"? The family $\Sigma$ need not be a topology, as union of simplices need not be simplices. $\endgroup$ – guidoar Mar 11 '20 at 3:24
  • $\begingroup$ Oh right... You're 100% correct. @Guido $\endgroup$ – Noel Lundström Mar 11 '20 at 19:57
  • $\begingroup$ I'm trying to come up with a functor $F:\text{SimpComplex} \rightarrow \textbf{Top}$ such that the underlying set of points of $FV$ is the set of vertices of $V$ and such that $FV$ has the same homotopy groups as $\mid V \mid$ but maybe this isn't as easy as I thought? $\endgroup$ – Noel Lundström Mar 11 '20 at 20:21
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    $\begingroup$ @NoelLundström: No such functor exists. For instance, there is a simplicial complex with three vertices with nontrivial $\pi_1$, but there is no space with three points with nontrivial $\pi_1$. There is a functor almost like that which instead sends a simplicial complex to a space whose underlying set is its set of simplices (with the topology corresponding to the inclusion order on the simplices). $\endgroup$ – Eric Wofsey Mar 11 '20 at 20:41
  • $\begingroup$ Where can I find more information on this functor you are describing? Sounds interesting... Is there something similar for a simplicial set $X$ aswell? Maybe define an ordering on $\sqcup_i X_i$ with $X_m \ni \phi \leq \sigma \in X_n$ if there exists a $\theta: [m] \rightarrow [n]$ with $X\theta (\sigma) = \phi$? $\endgroup$ – Noel Lundström Mar 11 '20 at 20:54
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I think that the confusion comes from two different meanings of the term "simplicial complex".

The pair $X=(V,\sum_V)$ that you've defined is typically called an abstract simplicial complex and in general it is not a topological space. The only case when $(V,\sum_V)$ is a topological space is when $\sum_V=2^V$ because that's the only situation when $V\in\sum_V$.

But indeed, every abstract simplicial complex $X$ induces a geometric realization $|X|$ which is a (concrete) simplicial complex. And every abstract simplicial function $f:X\to Y$ (i.e. $f$ maps $\sum_X$ to $\sum_Y$) indeed induces a continuous simplicial map $|f|:|X|\to|Y|$. It is not hard to see that not every continuous map $g:|X|\to|Y|$ arises in that way. Even up to homotopy, although up to homotopy $g$ can be approximated by simplicial maps.

I can't 100% understand the answer because he talks about simplicial complexes being homeomorphic to the nerve of some poset, which is a simplicial set?

The nerve of some poset is indeed a (concrete) simplicial complex. For a given poset $P$ you associate $n$-simplex with every sequence $a_1<a_2<\cdots<a_n$ in $P$. This is a topological space and so we can talk about homeomorphism. And this is the context of the answer you referred to.

There is an abstract version as well. If $P$ is a poset then we can associate an abstract simplicial complex with it by taking

$$\sum\nolimits_P=\big\{\{a_1,\ldots,a_n\}\subseteq P\ |\ a_1<\cdots <a_n\big\}$$ $$A(P)=\big(P, \sum\nolimits_P\big)$$

In that situation the nerve of $P$ is simply $|A(P)|$. But let's stay in the abstract context. We can ask whether any abstract simplicial complex $X$ is isomorphic to $A(P)$. By "isomorphism" we understand an abstract simplicial map $f:X\to A(P)$ which has an abstract simplicial inverse. But this doesn't touch any notion of topology.

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