Homology module of projective resolution

Question

I do not understand our first example for a homology module.

Let $A, B$ be $R$-modules ($A$ a left one, $B$ a right one), $0 \to P_1 \to P_0 \to A \to 0$ a projective resolution (exact).

Then consider $P_1 \to P_0 \to A \to 0$ exact. Since ${} \otimes B$ is a right exact functor, the sequence

$P_1 \otimes B \to P_0 \otimes B \to A \otimes B \to 0$

should be exact, right? (supplement with $0 \to P_1 \otimes B$ would not be exact at $P_1 \otimes B$)

Therefore (splicing)

$P_1 \otimes B \to P_0 \otimes B \to 0$

should be exact too, right?

The lecture notes (dealing with the homology of Tor) say $H_0(A) = A \otimes B$. But why?

We consider the chain map

$\dots \to 0 \to P_1 \to P_0 \to 0 \to \dots$

Thus \begin{align} H_0(A) & = Ker(P_0 \to 0) / Im(P_1 \to P_0) \\ & = P_0 / P_0 \\ & = \{0\} \end{align}

(In congruence with exactness at $A$ $\implies H_n(A) = 0$)

What did I understand wrongly? It drives me crazy that I do not understand this simple fact.

Answer 1

$P_1 \otimes B \to P_0 \otimes B \to A \otimes B \to 0$ is indeed exact because $\otimes B$ is a right exact functor. This does not however imply that $P_1 \otimes B \to P_0 \otimes B \to 0$ is exact. For example, consider the exact sequence $$\mathbb{Z} \xrightarrow{2} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.$$ Now if you drop $\mathbb{Z}/2\mathbb{Z}$ from it, you get $$\mathbb{Z} \xrightarrow{2} \mathbb{Z} \to 0$$ which is not exact.

Let's get back to the projective resolution of $A$ $$0 \to P_1 \xrightarrow{d_1} P_0 \xrightarrow{\epsilon} A \to 0$$ which gives you a right exact sequence $$ P_1 \otimes B \xrightarrow{d_1 \otimes 1_B} P_0 \otimes B \xrightarrow{\epsilon \otimes 1_B} A \otimes B \to 0,$$ so we have $\mathrm{Im}(d_1 \otimes 1_B) = \mathrm{Ker}(\epsilon \otimes 1_B)$.

For the computation of $\mathrm{Tor}$ we consider the deleted projective resolution $$0 \to P_1 \xrightarrow{d_1} P_0 \to 0,$$ which is no longer exact. And then tensoring it with $\otimes B$ gives you the sequence

$$ 0 \to P_1 \otimes B \xrightarrow{d_1 \otimes 1_B} P_0 \otimes B \to 0.$$

Now $\mathrm{Tor}_n(A,B) = \frac{\mathrm{Ker}(d_n \otimes 1_B)}{\mathrm{Im}(d_{n+1} \otimes 1_B)}$, where $d_0 \otimes 1_B$ is understood to be the zero map. So $$\mathrm{Tor}_0(A,B) = \frac{\mathrm{Ker}(d_0 \otimes 1_B)}{\mathrm{Im}(d_{1} \otimes 1_B)} = \frac{P_0 \otimes B}{\mathrm{Ker}(\epsilon \otimes 1_B)} = A \otimes B $$ and $$ \mathrm{Tor}_1(A,B) = \frac{\mathrm{Ker}(d_1 \otimes 1_B)}{\mathrm{Im}(d_{2} \otimes 1_B)} = \mathrm{Ker}(d_1 \otimes 1_B), $$

where $d_2 \otimes 1_B$ is again understood to the zero map.