With apologies to fellow algebraic topologists, I confess that I have no idea how to answer this innocent-looking question:



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> (**1**) Let's say we know that a finite simplicial complex $S$ is the barycentric subdivision of some other simplicial complex $K$. Can we find $K$ (up to simplicial isomorphism) from knowledge of $S$ alone? 
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Nor can I find answers to the related question:



> 
> (**2**) Given a simplicial complex $S$, can we decide if an "inverse-subdivision" $K$ exists? That is, can we decide if $\text{Sd}K = S$ has a solution $K$?
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Ideally, one would like an affirmative answer to:



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> (**3**) Are there efficient algorithms which decide (**2**) and solve (**1**), given $S$?
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Efficient in this case would mean polynomial time in the number of simplices in $S$ (this might be super-exponential in the number of simplices in $K$, if $K$ exists).




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**Background:**


Let $K$ be an abstract simplicial complex. Its *barycentric subdivision* $\text{Sd}K$ is defined to be the simplicial complex whose vertices are the simplices $\sigma \in K$, and each $d$-simplex is a sequence $\sigma\_0 < \sigma\_1 < \cdots < \sigma\_d$ of strict face relations in $K$. 


It is well-known that the geometric realizations of $\text{Sd}K$ and $K$ are homeomorphic, and barycentric subdivisions are used with remarkable frequency in basic algebraic topology (e.g., in proving the equivalence of simplicial and singular homology, or that simplicial homology satisfies the excision axiom, see Hatcher Ch. 2.1).




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**Update:**


From the comments, there doesn't even appear to be a consensus on whether non-isomorphic simplicial complexes can have isomorphic barycentric subdivisions. Maybe someone wants to take a stab at this foundational question as well.



Can we invert barycentric subdivision?
[As requested by Vidit Nanda, I am reposting a slightly edited version of my comment above as an answer. Nevertheless, I hope someone will eventually give a satisfactory answer to this question.]


The following article is relevant for this question: "[*The insufficiency of barycentric subdivision*](http://projecteuclid.org/euclid.mmj/1028999363)" by Ross L. Finney (published in Michigan Mathematical Journal, Volume 12, Issue 3 (1965), pages 263-272).


Let $K$ and $L$ be locally finite simplicial complexes. The cited article claims that $K$ and $L$ are isomorphic if their barycentric subdivisions are isomorphic. It also states that any isomorphism between the barycentric subdivisions of $K$ and $L$ is induced by an isomorphism $K\to L$, as long as $K$ is connected and verifies the following two conditions:


* $K$ is not isomorphic to a simplex or to the boundary of a simplex, and
* the geometric realization of $K$ is not a $1$-manifold without boundary (i.e. not homeomorphic to $S^1$ or to $\mathbb{R}$).


Perhaps the analysis leading to this latter result may help with problem 3 in the question.


