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?
 A: Let $f,g: X\to Y$ be homotopic maps. Then the corresponding chain maps satisfy
$$f_{\#}-g_\# = sd+ds. $$
As you already noticed, the right side, i.e. $sd +ds$, maps cycles to boundaries. Thus it is zero on homology, (which is exactly the point of this construction), so $$f_*=g_*.$$
In your case, $f$ is null-homotopic, so $g$ is a constant map.
Your confusion comes, because you think that if $g$ is constant, then it is zero on homology. But this is not true.
As $H_0$ corresponds to the path components, $g_0$ cannot be zero. If both spaces are path connected, then $g_0=\operatorname{Id}$, and otherwise it is the projection on the factor coming from the path-component in which the image of $g$ is conteined.
It is however true, that $g_i=0$ for $i>0$; Because if $g$ is constant, then it factors through a singleton, so $$g_i: H_i(X) \to H_i(\{x_0\}) \to H_i(Y),$$ and $H_i(\{x_0\})=0$ for $i>0$.
A: 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. 
