I think I have an intuitive understanding of why the following statement might be true, but I am not sure how to go about proving it. It's also possible my intuitive understanding is wrong and the statement is not true.

Let $G$ be a finite connected directed graph. Let $G_1, G_2, \ldots$ be a sequence of non-empty subgraphs of $G$ and let $f_i\colon G_{i+1}\to G_i$ be a sequence of simplicial maps respecting orientation of edges (for my purposes I also assume that $f_i$ does not act non-degenerately on any edges, so the image of an edge is always another edge - I don't know if this is necessary but I mention it anyway).


The topological inverse limit $\displaystyle{\lim_{\longleftarrow}(G_i,f_i)}$ is also a finite graph.

My intuition seems to come from the case of finite sets where a sequence of functions between subsets of some finite set $X$ will have an inverse limit which has cardinality less then or equal to $X$.

The maps $f_i$ in my setting are induced by functions acting on the finite set of edges in $G_i$ so you would expect something similar to happen in this setting, though I can't be sure. Is it as simple as parametrising points in $G_i$ with points in $E\times [0,1]$ where $E=\{\mbox{edges in } G_i\}$ (noting that $G_i\cong (E\times [0,1])/{\sim}$ where $\sim$ identifies ends of edges when they meet at a vertex) and then because $E$ is a finite set we can use the result from the case of finite sets? It's possible there's a category theory result which can be leveraged, though I'm not much of a category theorist so I don't know how that would look.

I should add, I normally study these inverse limits where the maps are not necessarily simplicial, and relaxing that condition can cause a lot of very interesting things to happen (you get spaces which look like solenoids and worse) but needless to say, the simplicial condition is crucial here, if indeed the claim is true.

  • $\begingroup$ I'm confused. By "topological inverse limit", do you mean inverse limit in $\mathbf{Top}$? In which case, are you identifying graphs with their geometric realisations? $\endgroup$
    – Zhen Lin
    Oct 16, 2014 at 22:03
  • $\begingroup$ @ZhenLin Yes and yes, and with all edges having unit length and $f_i$ mapping edges isomorphically onto other edges. Sorry I should have made it clear that I'm viewing the graphs as metric spaces. $\endgroup$
    – Dan Rust
    Oct 16, 2014 at 22:18
  • $\begingroup$ Hmmm. Let $G_n$ be the graph with vertices $\{ 0, \ldots, n \}$ and one edge connecting $0$ to each $i > 0$. Define maps $f_n : G_{n+1} \to G_n$ by collapsing $n$ and $n + 1$; then the inverse limit has infinitely many vertices. Does this fit your criteria? $\endgroup$
    – Zhen Lin
    Oct 16, 2014 at 22:38
  • $\begingroup$ @ZhenLin All the $G_n$ have to be subgraphs of a single finite connected graph $G$. $\endgroup$
    – Dan Rust
    Oct 17, 2014 at 14:11

1 Answer 1


Let $V_i$ and $G_i$ be the vertex and edge set of $G_i$ respectively. Note that $V_i\subset V(G)$ and $E_i\subset E(G)$ where $V(G)$ and $E(G)$ are the vertex and edge sets of $G$ respectively. Let $v_i\colon V_{i+1}\to V_i$ and $e_i\colon E_i\to E_i$ be the maps induced by $f_i$ on these finite sets in the obvious way. It's a general property of finite sets that if $a_i\colon A_{i+1}\to A_i$ is a sequence of functions on the finite sets $A_i$ such that $A_i\subset A$ for all $i$ and some finite set $A$, then $A_{\infty}=\varprojlim{_{FinSet}}(A_i,a_i)$ is a finite set and $\mbox{card}(A_{\infty})\leq \mbox{card}(A)$.

The finite set $V_{\infty}$ will form our vertex set and $E_{\infty}$ will form our edge set. We now need to prove that the edges we get can be assigned between the vertices in a way which corresponds with the topological inverse limit.

Because our graphs are directed and our simplicial maps $f_i$ respect orientation, we have distinguished (surjective) maps $s_i\colon E_i\to V_i$ and $t_i\colon E_i\to V_i$, the source and target maps, which send an edge to its source and target vertex respectively and we get commuting squares $$s_i\circ e_i = v_i\circ s_{i+1}\\ r_i\circ e_i = v_i\circ r_{i+1}$$ which induce (surjective) maps $s\colon E_{\infty}\to V_{\infty}$ and $t\colon E_{\infty}\to V_{\infty}$ that uniquely determine the source and target vertex of edges appearing in $E_{\infty}$. Because all of these maps were induced by the simplicial maps $f_i$, I claim without proof that the geometric realistion of the graph $G_{\infty}=(V_{\infty},E_{\infty})$ is homeomorphic to the inverse limit of the maps $f_i$.


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