Why is the induced orientation of a simplex well-defined?

Question

In my algebraic topology class I learned that let $A = (v_0,\cdots, v_k)$ be an oriented simplex and $B$ the face of $A$ by removing the vertex $v_i$, then the orientation for $A$ induces an orientation for $B$, namely $$ (-1)^i (v_0,\cdots, \hat{v_i}, \cdots, v_k). $$ My question is, why is the induced orientation well-defined? Since an orientation is an equivalence class, if $(w_0,\cdots, w_k)$ differs $(v_0,\cdots, v_k)$ by an even orientation, how can we make sure that $(w_0,\cdots, w_k)$ induces the same orientation for $B$?

Answer 1

Since every even permutation is a product of an even number of adjacent transpositions, it suffices to prove that the orientation of $B$ is switched if you do a single adjacent transformation on $A$, where the order of a single adjacent pair of entries $v_m,v_{m+1}$ is transposed (for some $1 \le m < m+1 \le k$).

One now does a case-by-case analysis.

If $1 \le m < m+1 \le i-1$, or if $i+1 \le m < m+1 \le k$, then the transposition on $A$ restricts to a transposition on $B$ and so the orientaiton on $B$ is indeed switched.

If $i-1=m < m+1=i$, or if $i=m < m+1=i+1$, then the position of the vertex $v_i$ in $B$ is shifted by $1$ unit before $v_i$ is removed. The exponent $i$ on the factor $(-1)^i$ is therefore also shifted by $1$ unit, and so the orientation on $B$ is also switched.