$\DeclareMathOperator{\ZZ}{\mathbb{Z}}$
$\DeclareMathOperator{\SS}{\mathbb{S}}$
$\DeclareMathOperator{\RP}{\mathbb{RP}}$
Let $R$ be a ring, $M$ an $R$-module,
A projective resolution of $M$ is an exact sequence of $R$-modules
$$ \cdots \to C_2 \xrightarrow{\partial_2} C_1 \xrightarrow{\partial_1} C_0 \xrightarrow{\varepsilon} M \to 0$$such that each $C_i$ is a projective $R$-module.
Let $N$ be another $R$-module.
Tensoring $N$ with $C_\bullet$, we get another chain complex
$$ \cdots \to N \otimes C_2 \to N \otimes C_1\to N \otimes C_0 \to 0.$$Taking the homology of this complex yields:
$$ \text{Tor}^R_n(N,M) := H_n(N \otimes_R C_{\bullet})$$Let $G$ be a (discrete) group. Want to do the above with the integral group ring
$$R = \mathbb{Z}G$$and its modules.
The group homology of $G$ is $$ H_{\bullet}G := \text{Tor}^{\mathbb{Z}G}_{\bullet}(\ZZ,\ZZ)$$
which is the homology of the chain complex
$$ \cdots \to \ZZ \otimes_{\mathbb{Z}G} C_2 \to \ZZ \otimes_{\mathbb{Z}G} C_1 \to \ZZ \otimes_{\mathbb{Z}G} C_0 \to 0$$where
$$\cdots \to C_2 \to C_1 \to C_0 \to \mathbb{Z} \to 0$$is a projective resolution of $\mathbb{Z}$.
How do we get projective resolutions of $M = \mathbb{Z}$?
Why do we tensor with $N = \mathbb{Z}$?
Topology has the answers!
Proposition. Let $G$ act freely on a space $E$.
Then $S_n E$ is a free $G$-set and $C_n E = \mathbb{Z}S_n E$ is a free $\mathbb{Z}G$-module.
Lemma. If $S$ is a free $G$-set, then $\mathbb{Z}S$ is a free $\mathbb{Z}G$-module.
So we get a chain complex of free $G$-modules
$$ \cdots \to C_2 E \to C_1 E \to C_0 E \xrightarrow{\varepsilon} \mathbb{Z} \to 0$$and if $E$ is contractible, $\tilde{H}_\bullet E = 0$, so this is exact.
We get a free resolution of $\mathbb{Z}$!
Let $M$ be a $\mathbb{Z}G$-module. The co-invariants of $M$ are
$$M_G := M/\langle gm - m\rangle.$$Lemma. For $S$ an arbitrary $G$-set,
$$\left(\mathbb{Z}S\right)_G \cong \mathbb{Z}(S/G).$$Proposition. Let $X$ be a space with $\pi_1(X) = G$ and contractible universal cover $E$. Then
$$(S_n E)/G \cong S_n (E/G),$$and thus,
$$(C_\bullet E)_G \cong C_\bullet X.$$Proposition. $$M_G \cong \mathbb{Z} \otimes_{\mathbb{Z}G} M.$$
Putting it together...
Let $X$ be a path-connected space with $\pi_1(X) = G$ and contractible universal cover $E$.
Then $G$ acts freely on $E$, so we get a free resolution of $\ZZ$:
$$ \cdots \to C_1 E \to C_0 E \xrightarrow{\varepsilon} \ZZ \to 0,$$Tensoring $C_\bullet E$ with $\ZZ$ yields
$$\cdots \to \ZZ \otimes C_1 E \to \ZZ \otimes C_0 E \to 0,$$which is isomorphic to
$$\cdots \to C_1 X \to C_0 X \to 0.$$Taking homology, $$H_\bullet X \cong H_\bullet G.$$
$X$ is a $K(G,1)$-space (complex) if:
Theorem. The homotopy type of a $K(G,1)$-complex is uniquely determined by $G$.
We could have thus defined:
$$H_\bullet G :=\, H_\bullet\, K(G,1)$$Proposition. [Hatcher, 1B.9]
Let $X$ be a connected CW-complex, $Y$ a $K(G,1)$-complex. Then every
$$\varphi: \pi_1(X,x_0) \to \pi_1(Y,y_0)$$is induced by
$$f: (X,x_0) \to (Y,y_0),$$and $f$ is unique up to homotopy (relative to $x_0$).
In fact, $K(G,1)$-complexes exist!
K(G,1)")