Let $X=\{(z_1,z_2) : z_1 , z_2 \in \mathbb{C}$ $and$ $ z_1 \neq z_2\}$. Compute the homology group of $X$. $\\$
My try: The space is nothing but $\mathbb{C} \times \mathbb{C} $ subtracting the diagonal. So some kind of excision theorem may be applied. But I have no clue. Please help me.
Define $F:X\times I\rightarrow X$, $(x,t)\mapsto F_t(x)$, by
$$F_t(z_1,z_2)=((1-t)z_1,(1-t)z_2-t(z_1-z_2))$$
and check that it is well-defined on $X$. Note that $F_0=id_X$ whilst $F_1(z_1,z_2)=(0,z_2-z_1)$. Moreover we have that $F_t(0,z_2)=(0,z_2)$, $\forall t\in I$, $\forall z_2\in \mathbb{C}\setminus\{0\}$, so $F$ is a deformation retraction onto the subspace $Y:=\{0\}\times \mathbb{C}\setminus\{(0,0)\}$. In turn $Y$ deformation retracts onto the unit circle $S^1\subseteq Y$. I'll leave you to write this one down.
Hence $X\simeq S^1$, and hopefully we all know the homology of this space.