# Connectedness of combinatorial complexes with no free faces

I'm currently reading the paper "$\mathcal{VH}$ complexes, towers and subgroups of $F \times F$" by Bridson & Wise. There they define combinatorial complexes as follows: A continous map between CW complexes is said to be combinatorial if it sends open cells homeomorphically onto open cells. A combinatorial complex is a CW complex all of whose attaching maps are combinatorial. About free faces they say: A cell e of a combinatorial complex X is called a free face if it lies in the boundary of exactly one cell e' of higher dimension and the intersection of the interior of e' with a small neighbourhood of an interior point of e is connected (the second clause in the preceding sentence is necessary in order to avoid suggesting, for example, that if both ends of a 1-cell are attached to a single vertex in a graph then that vertex is a free face).

Now my question: Let $\mathbb{K}$ be a class of (not necessarily compact) combinatorial complexes that is closed under the operations of passing to finite subcomplexes and to connected covers. If $Z \in \mathbb{K}$ compact and of dimension at most 2, with no free faces, is Z then connected or path-connected (equivalent in CW complexes) and if yes why? Thanks in advance

No, this is false. Let $Z = T \cup T'$ be the disjoint union of two toruses. Let $X$ be obtained from the union of $Z$ with an annulus $A$ by gluing one component of $\partial A$ glued to a circle in $T$ and the other component of $\partial A$ to a circle in $T'$. Then $X$ is a combinatorial complex, and $Z$ is a subcomplex of $X$ with no free faces, but $Z$ is not connected.
Edit: To address the question in the comments, here are a few more details. Take $\mathbb{K}$ to be the closure of the complex $X$ under passing to subcomplexes and to connected covers. So $Z \in \mathbb{K}$ because it is a subcomplex of $X$. And $Z$ is compact, of dimension 2, and has no free faces. But $Z$ is disconnected.