# Transfinite sequence of contiguous simplicial maps

Recall that two simplicial maps of (abstract) simplicial complexes $f,g\colon K\to L$ are contiguous if $f(\sigma)\cup g(\sigma)$ is a simplex of $L$ for every simplex $\sigma\in K$. Contiguous simplicial maps have homotopic geometric realisations: $|f|\simeq |g|$.

Let $K$, $L$ be simplicial complexes and let $(f_\phi\colon K\to L)_{\phi\leq \alpha}$ be a transfinite sequence of simplicial maps such that:

1. $f_\phi$ is contiguous to $f_{\phi+1}$ for all $\phi<\alpha$;
2. for each limit ordinal $\psi$ and each vertex $v\in K$ there is some $\beta<\psi$ such that $f_\phi(v)=f_\beta(v)=f_\psi(v)$ for each $\beta\leq \phi<\psi$.

Is there a homotopy between $|f_0|$ and $|f_\alpha|$?

If $\alpha$ is countable, we can embed it into $[0,1]$ and construct a homotopy by gluing homotopies between consecutive maps in the sequence. (I believe this should be continuous.) But what happens for arbitrary $\alpha$?

Is there some general theorem that answers this kind of questions? For a sequence of spaces instead of a sequence of maps I would use the Whitehead theorem, but see Whitehead for maps.

For each vertex $v$ and each $\beta\leq\alpha$, one can show by induction on $\beta$ that $\{f_\phi(v):\phi\leq\beta\}$ is finite. Using this, we see that for each finite subcomplex $F\subset K$, the restrictions $f_0|_F$ and $f_\alpha|_F$ are contiguous in the usual sense and therefore homotopic. However, this is not in general enough to conclude that $f_0$ and $f_\alpha$ are homotopic. In the case where $|L|$ is a homotopy commutative $H$-space and $K$ is the union of a chain of finite based subcomplexes $F_n$, we have a Milnor exact sequence $$0 \to \lim{}^1_n [\Sigma |F_n|,|L|] \to [|K|,|L|] \to \lim{}_n [|F_n|,|L|] \to 0$$ If I remember correctly, the $\lim^1$ term is nonzero (and very large) when $|K|=\mathbb{C}P^\infty$ and $|L|=S^3$. It is not clear to me whether one can choose triangulations and simplicial approximations in such a way as to convert this to a counterexample to your original question, but it seems like the right place to look.