# Nerve theorem for locally infinite covers by subcomplexes

Let $$Y$$ be a simplicial complex and let $$\{Y_i\}_{i\in I}$$ be a set of subcomplexes of $$Y$$ such that $$\bigcup_{i\in I}Y_i=Y$$. Let $$\mathcal N$$ be the nerve of this covering, and assume that for each finite $$J\subset I$$, we have that $$\bigcap_{j\in J}Y_j$$ is either empty or contractible.

One version of the nerve theorem says that, in the above situation, $$Y$$ is homotopy equivalent to $$\mathcal N$$.

I'm interested in the proof in the specific situation where we cannot assume that each simplex of $$Y$$ is contained in finitely many of the $$Y_i$$.

The theorem is stated in the above generality in , as Theorem 10.6. In the proof given there, it is assumed "for convenience" that the preceding local finiteness assumption holds, and that in the general case one uses a "slightly different" argument.

I tried a bit to deduce it from other versions of the nerve theorem (see e.g. ), say by trying to replace $$\{Y_i\}$$ by an open covering with a similar nerve. (I had some difficulty finding a more detailed proof in the literature, and I would like to see one to see if it's possible to modify it slightly.)

What is the "slightly different" argument mentioned in , or where can it be found?

: Björner, Anders, Topological methods, Graham, R. L. (ed.) et al., Handbook of combinatorics. Vol. 1-2. Amsterdam: Elsevier (North-Holland). 1819-1872 (1995). ZBL0851.52016.

: Proposition 4G.2 and Corollary 4G.3 in:

Hatcher, Allen, Algebraic topology, Cambridge: Cambridge University Press (ISBN 0-521-79540-0/pbk). xii, 544 p. (2002). ZBL1044.55001.