Question about orientations and loops of the $n-1$ faces of an n-simplex $\Delta^n$ $S$ is the $\Delta$-complex formed by considering $\Delta^n$ along with its faces. Let the vertices be $v_0, v_1, ... v_n$.
Then $C_n(S) = \langle[v_0 v_1..v_n]\rangle$
and $C_{n-1}(S) = \langle\{[v_0....\hat{v_i} ..v_n] : i \leq n \} \rangle$.
Now obviously $\partial_{n-1}(\Sigma_{i\leq n} (-1)^i[v_0..\hat{v_i}..v_n]) = 0$
Suppose I have $X \subseteq \partial\Delta^n$ a $\Delta$ complex. I want to show that if $H_{n-1}(X) \not = 0$, then $X = \partial \Delta^n$, but this amounts to showing that there can't be a cycle consisting of $< n+1$ many faces of the generators of $C_{n-1}(S)$. That is, I have to show that if $\partial_{n-1}(\Sigma_{i\leq m}N_i\times[v_0..\hat{v_i}..v_n]) = 0$, where $N_{i}$ are non-zero integers,then $m = n$. Thus $X$ contains all $n+1$ faces and then $X = \partial\Delta^n$.
This is intuitively pretty obvious, but I'm not sure how to rigorously show this. Hopefully this isn't too dumb of a question.
Is it obvious as saying that then $\Sigma_{i\leq m}N_i\times[v_0..\hat{v_i}..v_n]$ would then be in $\partial\Delta^n$'s $ker(\partial_{n-1})$, but then that woud be pushing back the question to showing that for $\partial\Delta^n$, $ker(\partial_{n-1}) = \langle\Sigma_{i\leq n}(-1)^i[v_0..\hat{v_i}..v_n]\rangle$, which, again is very obvious, but I'm blanking how to rigorously prove it.
Perhaps I'm missing something incredibly obvious however.

The $(n-1)$-st homology of your set $X \subset \partial \Delta^n$ is the same as the $(n-1)$-st homology of the union of the (closed) $(n-1)$-dimensional faces of $X$. So we can assume that $X$ is just a union of $(n-1)$-dimensional faces. If $X \not= \partial \Delta^n$, then roughly $X \subset \partial \Delta^n$ looks like the complement of an $(n-1)$-ball in the $(n-1)$-sphere $S^{n-1}$, which is contractible.
Even if it's hard to see why $X$ looks like the complement of a ball, it's at least the complement of a finite set of open balls in $S^{n-1}$. And such a thing deformation retracts to an $(n-2)$-complex.
