TITLE: Connecting homomorphism in Galois homology using the standard resolution
QUESTION [1 upvotes]: Let $G$ be a finite group, although this may not be necessary for almost everything that follows. One of the ways of defining Galois homology groups is using the standard resolution for the $\mathbb{Z}$, explicitly we do the following. Let 
$$ \cdots\to P_2\to P_1 \to P_0 \to \mathbb{Z} $$ 
be the standard resolution. Let $I\subset \mathbb{Z}[G]$ be the ideal generated by elements of the type $(g-1)$. For two $G$-modules $M,N$ the $G$ module structure on $M\otimes N$ is given by $$g\cdot (m\otimes n):=gm\otimes gn.$$
Define 
$$H_i(G,M)=H_i\bigg(\cdots\to \frac{P_2\otimes M}{I(P_2\otimes M)}\to \frac{P_1\otimes M}{I(P_1\otimes M)} \to \frac{P_0\otimes M}{I(P_0\otimes M)}\to 0\bigg).$$
If $$0\to M'\to M\to M''\to 0$$ is a short exact sequence of $G$-modules then we have a short exact sequence of $G$-modules 
$$0\to \frac{P_i\otimes M'}{I(P_i\otimes M')}\to \frac{P_i\otimes M}{I(P_i\otimes M)} \to \frac{P_i\otimes M''}{I(P_i\otimes M'')}\to 0.$$
As a consequence we get a short exact sequence of complexes 
$$0\to \frac{P_\bullet\otimes M'}{I(P_\bullet\otimes M')}\to \frac{P_\bullet\otimes M}{I(P_\bullet\otimes M)} \to \frac{P_\bullet\otimes M''}{I(P_\bullet\otimes M'')}\to 0$$
and consequently a long exact sequence of homology groups. The connecting homomorphism is defined in the usual way; we start with an element in $\frac{P_i\otimes M''}{I(P_i\otimes M'')}$ which is 0 under the differential, lift it to an element in $\frac{P_i\otimes M}{I(P_i\otimes M)}$ and apply the differential to get an element in $\frac{P_{i-1}\otimes M'}{I(P_{i-1}\otimes M')}$, whose homology class is the required element. 
Now consider the case when $M=\mathbb{Z}[G]$ and $M''=\mathbb{Z}$, and $\bullet=1$. In this case the connecting homomorphism should give an isomorphism $H_1(G,\mathbb{Z})\to H_0(G,I)$.
Let $\alpha$ be an element in $\frac{P_1}{IP_1}$ whose differential is 0. We can lift this to the element $\alpha\otimes 1$ in $\frac{P_1\otimes \mathbb{Z}[G]}{I(P_1\otimes \mathbb{Z}[G])}$, but now applying the differential, which in this case is $d\otimes 1$, we will always get 0 since $d\alpha=0$, which is clearly not correct. The same problem would appear if $M''$ is a summand of $M$ as abelian groups. Could someone please point out what is it that I am doing wrong.

REPLY [1 votes]: What you're missing is that the equation $d\alpha=0$ is taking place in $P_0/IP_0$, not in $P_0$.  Backing up a bit, if $\alpha\in P_1/IP_1$ is a cycle, let's choose a lift $\beta\in P_1$ for $\alpha$.  Then you can't necessarily say that $d\beta=0$ as an element of $P_0$, since all you know is that the image of $d\beta$ in the quotient $P_0/IP_0$ is $0$, i.e. that $d\beta\in IP_0$.  Your lift of $\alpha$ to $P_1\otimes\mathbb{Z}[G]/I(P_1\otimes\mathbb{Z}[G])$ is then the coset of $\beta\otimes 1$, and applying the differential yields the coset of $d\beta\otimes 1$ in $P_0\otimes\mathbb{Z}[G]/I(P_0\otimes\mathbb{Z}[G])$.  But there's no reason for this to be $0$: you can't conclude from $d\beta\in IP_0$ that $d\beta\otimes 1\in I(P_0\otimes\mathbb{Z}[G])$, because $I$ acts on $P_0\otimes\mathbb{Z}[G]$ diagonally.
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