§ Zeroth singular homology group: Intuition

We wish to show that for a path connected space XX, the zeroth singular homology group is just Z\mathbb Z. The intuition is that the zeroth homology group is given by consider C[1]1C[0]C[1] \xrightarrow{\partial_1} C[0], C[0]00C[0] \xrightarrow{\partial_0} 0, and then taking H[0]ker(0)/im(1)=C[0]/im(1)H[0] \equiv ker(\partial_0) / im(\partial_1) = C[0] / im(\partial_1). Recall that C[0]C[0] is the abelian group generated by the direct sum of generators {Δ0X}\{ \Delta^0 \rightarrow X \}, where Δ0\Delta^0 is the 00-simplex, that is, a single point. So C[0]C[0] is an abelian group generated by all points in XX. Now, C[1]C[1] contains all paths between all points p,qXp, q \in X. Thus the boundary of C[1]C[1] will be of the form qpq - p. Quotienting by C[1]C[1] identifies all points with each other in C[0]C[0]. That is, we get H[0]xXy=zy,zXH[0] \equiv \langle x \in X | y = z \forall y, z \in X \rangle, which is isomorphic to Z\mathbb Z. Thus, the zeroth singular homology group is Z\mathbb Z.