Let $R$ be a ring and $M$ an $R$-module. Let $(x_{1},\ldots,x_{n})$ be a sequence of elements of $R$. Define the (homological) Koszul complex $K_{\bullet}((x_{1},\ldots,x_{n}),M)=:K_{\bullet}(n)$ to be given in degree $p\geqslant 0$ by the collection of maps $f\colon \{ 1,2,\ldots,n\}^{n-p} \to M$ with the following properties:
$f(i_{1},\ldots,i_{n-p})=0$ whenever $i_{k}=i_{l}$ for some $1\leqslant k < l \leqslant n-p $.
$f(i_{\pi(1)},\ldots, i_{\pi(n-p)})=\text{sgn}(\pi)f(i_{1},\ldots,i_{n-p})$ for every permutation $\pi \in S_{n-p}$.
The differential $d_{p}\colon K_{p}(n)\to K_{p-1}(n)$ is given by $$ (d_{p}f)(i_{1},\ldots,i_{n-p+1})=\sum_{j=0}^{n-p}(-1)^{j}x_{i_{j+1}}f(i_{1},\ldots,\hat{i_{j+1}},\ldots,i_{n-p+1}) $$
I am trying to understand this definition with exterior powers and their formal properties as follows:
A function $f\in K_{p}(n)$ is uniquely determined by giving the image of the tuples $(i_{1},\ldots,i_{n-p})$ for $1\leqslant i_{1} < \ldots < i_{n-p} \leqslant n-p$, and $K_{p}(n)$ is spanned additively by the set of functions which take some value $m\in M$ at one such tuple and zero in any other such tuple. Denote this function by $$ m(e_{i_{1}}\wedge \ldots \wedge e_{i_{n-p}}) $$
The usual properties of wedge products are then satisfied. Some examples:
For $n=0$ and $p=0$ we have functions $f\colon 0^{0}=\{ *\}\to M$. Such functions are the same as $M$ and we get the complex $$ K_{\bullet}(0): \quad M \leftarrow 0 \leftarrow \cdots $$
For $n=1$ and $p=1$ we have functions $f\colon \{1\}^{0}=\{ *\} \to M$. Again, this is just $M$. For $p=0$ we have functions $f\colon \{1\}\to M$, which is again just $M$. But now we have a differential: for $f\colon \{ *\}\to M$, we have $(d_{1}f)(1)=x_{1}f(*)$. Hence our complex is $$ K_{\bullet}(1): \quad M\xleftarrow{x_{1}} M \leftarrow 0 \leftarrow \cdots $$
For $n=2$ and $p=2$ we get $M$ as before. For $p=1$ we have maps $f\colon \{1,2\}\to M$, which are characterized by the images of $1$ and of $2$, so we get $M\oplus M$ with a basis $e_{1}$ and $e_{2}$. For $p=0$ we have maps $f\colon \{ 1,2\}^{2}\to M$ which are characterized by the image of $(1,2)$, so we get $M$ with a basis $e_{1}\wedge e_{2}$. For the differential we have $(d_{2}f)(1)=x_{1}f(*)$, $(d_{2}f)(2)=x_{2}f(*)$ and $(d_{1}g)(1,2)=x_{1}g(2)-x_{2}g(1)$. Hence our complex is $$ K_{\bullet}(2): \quad M\xleftarrow{\binom{-x_{2}}{x_{1}}} M\oplus M \xleftarrow{(x_{1},x_{2})} M \leftarrow 0 \leftarrow \cdots $$
Did I understand this definition correctly? Should I interpret the case $p=n$ the way I did?
Now comes the issue that confused me and made me suspect that I am getting something wrong: let $\varphi \colon K_{\bullet}(n-1)\xrightarrow{x_{n}} K_{\bullet}(n-1) $ and let $C_{\bullet}$ be its cone, with the following sign convention for the differential: \begin{align*} d_{i}^{C} \colon K_{i}(n-1)\oplus K_{i-1}(n-1) &\to K_{i-1}(n-1)\oplus K_{i-2}(n-1) \\ (x,y) & \mapsto (d_{i}x+x_{n}a,-d_{i-1}a) \end{align*}
For example, for $n=2$ we have a cone as follows
$$ M \xleftarrow{\binom{x_{1}}{x_{2}}} M\oplus M \xleftarrow{(x_{2},-x_{1})} M $$
Then the claim is that the following is a chain complex isomoprhism:
\begin{align*} K_{p}(n) & \to C_{p}=K_{p}(n-1)\oplus K_{p-1}(n-1) \\ f & \mapsto (f(n,-)|_{\{1,\ldots,n-1\}^{n-p-1}},-f|_{\{1,\ldots,n-1\}^{n-p}}) \end{align*}
In the case $n=2$, this is how I understood this map:
- For $p=2$, the left component is empty and the right component is just $-1$ (minus the identity), so we get
\begin{align*} K_{2}(2)=M & \to C_{2}=M \\ f(*)=m &\mapsto -f(*)=-m \end{align*}
- For $p=1$ we get
\begin{align*} K_{1}(2)=M\oplus M & \to C_{1}=M\oplus M \\ m_{1}e_{1}+m_{2}e_{2}=(m_{1},m_{2}) &\mapsto (f(2),-f(1))=(m_{2},-m_{1}) \end{align*}
- For $p=0$ we get
\begin{align*} K_{0}(2)=M & \to C_{1}=M \\ m(e_{1}\wedge e_{2}) &\mapsto f(2,1)-f(1,1)=-m-0=-m \end{align*}
So now, if $m\in M=K_{2}(2)$, then one composition is $m\mapsto -m \mapsto (-x_{2}m,x_{1}m)$, and the other one is $m\mapsto (x_{1}m,x_{2}m)\mapsto (x_{2}m,-x_{1}m)$.
Did I do something wrong? How can I understand this isomorphism (or the corrected version of this isomorphism) intuitively?