I am following Brown's 'Cohomology of groups' and the homology is defined as follows:
Let $\cdots \rightarrow F_{n}\rightarrow F_{n-1}\rightarrow \cdots \rightarrow F_{1}\rightarrow F_{0}\rightarrow \mathbb{Z}\rightarrow 0$ be a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$.
Apply the functor $\mathbb{Z} \otimes_{\mathbb{Z}G} -$, so we have $\cdots \rightarrow \mathbb{Z}\otimes_{\mathbb{Z}G}F_{n}\rightarrow \mathbb{Z}\otimes_{\mathbb{Z}G}F_{n-1}\rightarrow \cdots \rightarrow \mathbb{Z}\otimes_{\mathbb{Z}G}F_{1}\rightarrow \mathbb{Z}\otimes_{\mathbb{Z}G}F_{0}\rightarrow \mathbb{Z}\otimes_{\mathbb{Z}G}\mathbb{Z}\rightarrow 0$.
Then, $H_{i}G=H_{i}(F_{G})$.
I am not understanding something very basic: If projective resolutions are exact by definition and $\mathbb{Z} \otimes_{\mathbb{Z}G} -$ is a right-exact functor, why is not the homology 0?
Sorry for the stupid question and thanks in advance!
A right exact functor does not preserve arbitrary exact sequences, but only those of the form $A\to B\to C\to 0$.
More precisely, a right exact functor is by definition a functor that preserves finite colimits. We can build all finite colimits from cokernels and finite coproducts, so an additive functor (i.e. a functor that preserves finite products and coprpducts) is right exact if and only if it preserves cokernels, hence if and only if it preserves a sequence as the one above (because such a sequence is exact if and only if the map on the right is the cokernel of the map on the left).