Trouble understanding null-homotopic chain maps I'm working through Weibel, and I'm at the part where he defines null-homotopic maps. He says it's essentially topological null-homotopy, but I'm having trouble reconciling that with examples.

Remark: This terminology comes from topology via the following observation. A map $f$ between two topological spaces $X$ and $Y$ induces a map $f^*:  S(X) \to S(Y)$ between the corresponding singular chain complexes. It turns out that if $f$ is topologically null homotopic (resp. a homotopy equivalence), then the chain map $f^*$  is null homotopic (resp. a chain homotopy equivalence), and if two maps $f$  and $g$ are topologically homotopic, then $f^*$ and $g*$  are chain homotopic.

If $X$ is the circle and $Y$ is the sphere, and $f: X \to Y$ is the inclusion of the equator, then $f$ is topologically null-homotopic. The map it induces on the complex should decompose as $f^* = sd + ds$.
But I don't see how this can work in degree 0. Since $H_{-1}(X) = 0$ we must have $sd = 0$ for any choice of $s$. And similarly, $ds$ must be zero, because $H_1(Y) = 0$. But $f^*$ isn't zero, it's the identity.
What am I misinterpreting about this example?

It doesn't. $f$ induces an isomorphism on $H_0$ because $X$ and $Y$ are both path-connected, and this is true more generally for any map between path-connected spaces. In topology it's actually not possible for a map to induce zero on $H_0$ since the connected components of the source have to map to the connected components of the target somehow. If you want a tighter analogy to topological null-homotopy you should be looking at 1) pointed maps and 2) reduced homology. 
