Definition
An abstract simplicial complex $K$ is a pair $(X_K,\Phi_K)$ where $X_K$ is a set of points, and $\Phi_K\subseteq2^X$ such that:
- If $S\in\Phi_K$, then $2^S\subseteq\Phi_K$;
- $\{x\}\in\Phi_K$ for all $x\in X_K$.
An ordered abstract simplicial complex is an a.s.c. as defined above with $X_K$ having a total order.
Definition
$S=\{P_0,\dotsc,P_n\}\in\Phi_K$ (cardinality $n+1$) is an $n$-simplex of the complex $K$. We denote the set of $n$-simplices of $K$ as $K^{\Delta^n}$.
Definition
A simplicial map is a map $f:X_K\to X_L$ (where $K,L$ are a.s.c.'s) such that, for every $S\in\Phi_K$, $f(S)\in\Phi_L$, where, if $S=\{P_0,\dotsc,P_n\}$, we define $f(S):=\{f(P_0),\dotsc,f(P_n)\}$.
Note
$K^{\Delta^n}$ is precisely the set of simplicial maps from $\Delta^n$ to $K$.
Definition
For $n\in\mathbb{N}$, we define $C_n(K)$, with $K$ an o.a.s.c., to be the free abelian group generated by $K^{\Delta^n}$, i.e. $\mathbb{Z}[K^{\Delta^n}]=\{\sum_{i=1}^nn_ix_i:n_i\in\mathbb{Z},x_i\in K^{\Delta^n},n<\infty\}$.
Remark
We will always assume $X_K$ is finite, so the $n<\infty$ assumption above is only to make that OK in more general contexts as well as the present one. The above is a set of formal linear combination, that can be viewed as functions from $K^{\Delta^n}$ to $\mathbb{Z}$ sending each $x_i$ to the matching $n_i$, and the rest of $K^{\Delta^n}$ to 0.
Notation
If $\{P_0,\dotsc,P_n\}\in K^{\Delta^n}$, we will use square brackets to fix the order, i.e. $[P_0,\dotsc,P_n]$ as a chain is considered to have a sign which is the one of the permutation mapping $(P_0,\dotsc,P_n)$ to the ordered $n$-tuple with the same elements in the order given by $X_K$'s order. So assume $P_0\leq P_1\leq\dotso\leq P_n$, then $[P_1,P_0,\dotsc,P_n]=-[P_0,P_1,\dotsc,P_n]$, and, in general, $[P_{i_0},\dotsc,P_{i_n}]=\operatorname{sign}(i_0,\dotsc,i_n)[P_0,\dotsc,P_n]$. If $\{P_0,\dotsc,P_n\}$ is not an $n$-simplex, then either there are coinciding points, and it is a simplex of lower dimension, or it is not a simplex at all. In the former case, $[P_0,\dotsc,P_n]=0$. In the latter, $[P_0,\dotsc,P_n]$ will be left undefined.
Definition
If $f:K\to L$ is simplicial, we define, for each $n$, a group homomorphism $C_n(f)$ which acts on generators $\sigma\in K^{\Delta^n}$ as:
$$C_n(f)([\sigma])= \begin{cases} 0 & f(\sigma)\notin L^{\Delta^n} \\ [f(\sigma)] & f(\sigma)\in L^{\Delta^n} \end{cases},$$
and is extended to the rest of $C_n(K)$ by linearity, i.e.:
$$C_n(f)\left(\sum_in_ix_i\right)=\sum_in_iC_n(f)(x_i).$$
$[\sigma]$ and $[f(\sigma)]$ denote the simplices written in the square-bracket notation introduced above, hence with the sign adjustment.
Remark
I wanted to be explicit above, which is why I distinguished the cases, but naturally, the zero case is implicit in the notation.
Definition
For each $n$, we define $\partial_n:C_n(K)\to C_{n-1}(K)$, where $K$ is an o.a.s.c., as:
$$\partial_n([P_0,\dotsc,P_n])=\sum_i(-1)^j[P_0,\dotsc,\widehat{P_j},\dotsc,P_n],$$
plus extension by linearity as done for $C_n(f)$ above. The hat indicates the canceling of $P_j$, that is, $[P_0,\widehat{P_1},P_2]$, for example is the same as $[P_0,P_2]$. Hat notation comes in handy when you don't know which one you are removing. We call $\partial_n$ the $n$-th boundary operator.
Lemma
With the above definitions, we have, for each $n$, that:
$$C_n(f)\circ\partial_n^L=\partial_n^K\circ C_{n-1}(f).$$
Note that I put superscripts to distinguish the boundary operators of the two simplicial complexes.
This is basically all stuff we did in class. Unfortunately, the teacher left the proof of this lemma unfinished, I tried finishing it, and I got stuck. Here is my work, together with the unfinished proof given in class.
Proof.
First of all, if we verify this on simplexes, then we are done. There will be two cases.
(Case 1: $C_n(f)(\sigma)=0$, alias $f(\sigma)$ is not an $n$-simplex)
Obviously:
$$\partial_n^LC_n(f)(\sigma)=\partial_n^L(0)=0.$$
Let us see the other direction. First step is:
$$\partial\sigma=\sum_{i=0}^n(-1)^i[P_0,\dotsc,\widehat{P_i},\dotsc,P_n].$$
If there are at least two pairs of points which $f$ maps to the same point, then all the terms in that sum map to zero via $C_{n-1}(f)$ , hence the whole sum maps to zero, as we want. Else, we rewrite the sum as:
$$\partial\sigma=(-1)^j[P_0,\dotsc,\widehat{P_j},\dotsc,P_{j+k},\dotsc,P_n]+(-1)^{j+k}[P_0,\dotsc,P_j,\dotsc,\widehat{P_k},\dotsc,P_n]+\text{the rest}.$$
The rest contains only terms with both $P_j$ and $P_{j+k}$ left uncanceled, and those terms will map to zero, since they will have $f(P_j)=f(P_{j+k})$ both in the square brackets, which means the resulting chain is zero. So we need to show those two terms map to opposite terms which cancel out, giving us the zero we wish to obtain. To this end, we notice that:
$$[f(P_0),\dotsc,\widehat{f(P_j)},\dotsc,f(P_{j+k}),\dotsc,f(P_n)]=(-1)^{k-1}[f(P_0),\dotsc,\widehat{f(P_j)},f(P_{j+k}),\dotsc,\widehat{f(P_{j+k})},\dotsc,f(P_n)],$$
since between the canceled term and $f(P_{j+k})$ there are $f(P_{j+1}),\dotsc,f(P_{j+k-1})$, that is $k-1$ terms, and bringing $f(P_{j+k})$ one term back (i.e. swapping it with the term just before it) gives a minus sign to the permutation. But in the terms we must map via $C_{n-1}(f)$, we have $(-1)^j$ and $(-1)^{j+k}$, and implementing those, we get:
\begin{align*} (-1)^{j+k}[f(P_0),\dotsc,\widehat{f(P_j)},\dotsc,f(P_{j+k}),\dotsc,f(P_n)]={}&(-1)^{j+k+k-1}[f(P_0),\dotsc,\widehat{f(P_j)}, \\ &\hspace{1cm}f(P_{j+k}),\dotsc,\widehat{f(P_{j+k})},\dotsc,f(P_n)]{}={} \\ {}={}&(-1)^{j-1}[f(P_0),\dotsc,\widehat{f(P_j)}, \\ &\hspace{1cm}f(P_{j+k}),\dotsc,\widehat{f(P_{j+k})},\dotsc,f(P_n)]={} \\ {}={}&(-1)^{j-1}[f(P_0),\dotsc,f(P_j),\dotsc,\widehat{f(P_{j+k})},\dotsc,f(P_n)]={} \\ {}={}&-(-1)^j[f(P_0),\dotsc,f(P_j),\dotsc,\widehat{f(P_{j+k})},\dotsc,f(P_n)], \end{align*}
which is precisely what we wanted. Hence, case 1 is proved.
(Case 2: $f(\sigma)$ is an $n$-simplex)
In this case, we have:
\begin{align*} \partial_n^LC_n(f)(\sigma)={}&\partial_n^L[f(P_0),\dotsc,f(P_n)]=\sum_{j=0}^n(-1)^j[f(P_0),\dotsc,\widehat{f(P_j)},\dotsc,f(P_n)], \\ C_n(f)(\partial_n^K\sigma)={}&C_n(f)\left(\sum_{j=0}^n(-1)^j[P_0,\dotsc,\widehat{P_j},\dotsc,P_n]\right)=\sum_{j=0}^n(-1)^j[f(P_0),\dotsc,\widehat{f(P_j)},\dotsc,f(P_n)]. \end{align*}
Or do we? I believe the notation is cheating on us. What I mean is: in the first equality, we are expanding a boundary as if the points were already correctly ordered, but that was never an assumption! And this is precisely where I got stuck. $\hspace{2cm}\square$
Question
How do I fix that sign problem?
Note
In case 1, we were expanding the boundary before doing $C_n$, because the other composition was guaranteed to yield zero, so we could assume $P_0,\dotsc,P_n$ were correctly ordered. In this case, either we assume that, or that $f(P_0),\dotsc,f(P_n)$ are, so either (former assumption) equality 1 is potentially incorrect, or (latter assumption) equation 2 is.
Update
(Or rather, comment to dlf's post which I turn to an update because it might be too long for a comment)
You are right, I sort of forgot about order. Then again, if you think of it, that sign-adjusting bracket notation is basically implementing those relations :). Anyway, I originally tried (on my notebook) with the $(-1)^{\text{permutation}}$, and I still wasn't convinced the signs fell in place. With that permutation sign, equality one above becomes:
$$\partial_n^LC_n(f)(\sigma)=(-1)^\bullet\partial_n^L[f(P_0),\dotsc,f(P_n)]=\sum_{j=0}^n(-1)^\bullet(-1)^j[f(P_0),\dotsc,\widehat{f(P_j)},\dotsc,f(P_n)],$$
whereas equality two is:
$$C_n(f)(\partial_n^K\sigma)=C_n(f)\left(\sum_{j=0}^n(-1)^j[P_0,\dotsc,\widehat{P_j},\dotsc,P_n]\right)=\sum_{j=0}^n(-1)^j(-1)^{\bullet_j}[f(P_0),\dotsc,\widehat{f(P_j)},\dotsc,f(P_n)],$$
where the $\bullet$s are the signs of the permutations that were originally hidden inside the bracket notation. So the problem becomes:
$$\text{Why is }(-1)^{\bullet_j}(-1)^j=(-1)^\bullet(-1)^j?$$
For example, suppose the only permutation required is to swap $f(P_0)$ first with $f(P_1)$, then with $f(P_2)$, and finally with $f(P_3)$. That is 3 swaps, so $(-1)^\bullet=-1$, but the $j=0$ term in equality two would have $(-1)^{\bullet_j}=1$ since having removed $f(P_0)$ there is no permutation needed. So it seems I have a sign mismatch. Am I missing something?
Update
Trying to write things out in a simple case, I got the following.
Suppose we have $f(A)=3,f(B)=0,f(C)=1,f(D)=2$. Then:
\begin{align*} f([ABCD])={}&(-1)^\bullet[3012]=-[0123]; \\ \partial f([ABCD])={}&-\partial[0123]=-(123-023+013-012); \\ \partial[ABCD]={}&BCD-ACD+ABD-ABC; \\ f(\partial[ABCD])={}&(-1)^{\bullet_A}f(BCD)-(-1)^{\bullet_B}f(ACD)+(-1)^{\bullet_C}f(ABD)-(-1)^{\bullet_C}f(ABC)=012-312+302-301, \end{align*}
so the equality holds. But with the above expressions, we get:
\begin{align*} \partial f(ABCD)={}&\sum_{j=0}^3(-1)^\bullet(-1)^j[f(P_0),\dotsc,\widehat{f(P_j)},\dotsc,f(P_3)]={} \\ {}={}&-[(-1)^0[f(B),f(C),f(D)]+(-1)^1[f(A),f(C),f(D)]+(-1)^2[f(A),f(B),f(D)]+(-1)^3[f(A),f(B),f(C)]={} \\ {}={}&-[[012]-[312]+[302]-[301]], \end{align*}
with all signs wrong. The other expression gives the correct result, you can check. So I must have gotten something wrong in that expression… but what?
Said teacher here. Without stack-exchange reputation I cannot comment: this is not the answer, but a suggestion.
Actually, $C_n(K)$ is not the free $\mathbb{Z}$-module generated by simplicial maps $\Delta^n \to K$. Either order vertices and take only order-preserving simplicial maps, or quotient the module by adding all $(n+1)!$ relations of type $[P_0P_1\ldots P_n] + [P_1 P_0 \ldots P_n]=0$, to obtain $C_n(K)$.
Then, in defining $C_n(f)$ either put a $(-1)^\text{permutation}$ in front of $[f(s)]$ (for the ordered-vertices case) or use relations in the quotient module (which is what is used in the book, if I am not wrong). Then the signs will fall in place.
UPDATE:
Your example, if I am not wrong, is:
$A,B,C,D \to 1,2,3,0$.
Hence:
$\partial [A,B,C,D] = BCD-ACD+ABD-ABC$; (omitting commas and brackets)
$f(ABCD)=1230=-0123$;
$\partial f (ABCD) = - ( \partial [0123] ) = -123+023-013+012$;
$f(\partial (ABCD)) = f (BCD-ACD+ABD-ABC ) = 230 - 130 +120 - 123 = 023 -013 + 012 - 123$;
They coincide. Too fishy example?