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 [1], 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. [2]), 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 [1], or where can it be found?

[1]: *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.

[2]: 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.