Undecidability of simple-connectedness of a special class of simplicial complexes. It's a well known fact that determining simple-connectedness of a CW-complex is in general undecidable, but I'm interested whether that's also true in the following class of topological spaces.
Let $\Delta_n$ be a simplex with vertices $\{1, 2, \dots, n\}$, where we view $\Delta_n$ as a $(n - 1)$-dimensional simplicial complex. Now let $Y \subseteq \Delta_n$ be a $2$-dimensional simplicial subcomplex of $\Delta_n$, such that $Y$ contains all $0$-faces and all $1$-faces of $\Delta_n$ (that is, $Y$ contains the $1$-skeleton of $\Delta_n$).
Does there exists an algorithm that would in general determine whether $Y$ is simple-connected?
If not, what would be the counterexample? My first thought was to check whether every $3$-cycle of a complete graph $K_n$ is contractible in $Y$, but I don't know how you would do that (at least semi-efficiently).
 A: As suggested in the comment of @CheerfulParsnip, this can be algorithmically reduced to the problem of deciding whether the group given by a finite presentation is trivial, but the latter is an undecidable problem, hence your problem is also undecidable; that is to say, there is no such algorithm.
Here are the details of the algorithmic reduction.
Start from any finite presentation of a group $G$. Take the ordinary Cayley 2-complex and mod out by the action of $G$, to get a cell complex with one vertex having the given fundamental group $G$.
Next, take the 2nd barycentric subdivision of that cell complex, converting it into a simplicial complex having the same fundamental group $G$. At this stage, enumerating the vertices from $1$ to $n$, your simplicial complex is already embedded as a connected subcomplex $K$ of the 2-skeleton of $\Delta_n$, but it might not yet contain all of the edges.
By induction on the number of missing edges, given any missing edge $E \subset \Delta_n \setminus K$, I simply need to explain how to find a larger subcomplex $K'$ of the 2-skeleton such that $K \cup E \subset K'$ and such that $K'$ deformation retracts to $K$, and hence the inclusion map $K \hookrightarrow K'$ induces an isomorphism on fundamental groups.
Let $v,w$ be the endpoints of the edge $E$ that you want to add. Let $v=v_0,...,v_L=w$ be the shortest edge path in $K$ from $v$ to $w$ (such a path exists because $K$ is connected). Note that $L \ge 2$, because $E = \overline{v_0 v_L} \not\subset K$. More generally, since that path is shortest, for $0 \le i < i+2 \le j \le k$ the edge $\overline{v_i v_j}$ is not in $K$. What we add to $K$ in order to form $K'$ are some edges and 2-simplices, namely the edges
$$\overline{v_0 v_2} \quad \overline{v_0 v_3} \quad \ldots \quad \overline{v_0 v_L}
$$
and the 2-simplices
$$\overline{v_0 v_1 v_2} \quad \overline{v_0 v_2 v_3} \quad \ldots \quad \overline{v_0 v_{L-1} v_L}
$$
It follows easily that $K'$ deformation retracts to $K$.
