# 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.

• $s$ is only a map on chains, not on homology. So it does not neccesarily vanish. – klirk Dec 1 '17 at 18:24
• Ah, true. I think the issue persists though, $C_{-1}(X)$ is still zero, and $f^*$ can't equal $ds$ for any $s$, since the image of $f^*$ contains things that are not boundaries. – Henry Swanson Dec 1 '17 at 18:27
• Maybe you need to look at reduced homology and replace $d_0$ with $\epsilon$. – klirk Dec 1 '17 at 18:38
• Yeah, haven't worked out a proof, but reduced homology works with all the examples I've cooked up. A good argument for its importance :p – Henry Swanson Dec 1 '17 at 19:15

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.

• So the remark (in edit) is false, as written? That's a relief; I was starting to question what it means to be (topologically) null-homotopic... Thanks! – Henry Swanson Dec 1 '17 at 18:09
• Yes, the remark is false as written; there are a lot of mistakes in Weibel in general (sites.math.rutgers.edu/~weibel/Hbook.errors.edition2.pdf). The correct statement is just that if two maps $f, g : X \to Y$ are homotopic then the induced maps on chain complexes are chain homotopic. For spaces nullhomotopy means homotopic to the map which sends everything to a basepoint of $Y$, and for chain complexes nullhomotopy means homotopici to the map which sends everything to $0$. – Qiaochu Yuan Dec 1 '17 at 18:20
• I don't think I see why that resolves the problem with my example. It seems $f$ is homotopic to a constant map, and $f^*$, not being equal to $sd+ds$, isn't chain homotopic to the zero map. Is there a subtlety about "basepoint" that I'm missing? – Henry Swanson Dec 1 '17 at 18:32
• The basepoint of $Y$ generates $H_0$ rather than being zero there. – Qiaochu Yuan Dec 1 '17 at 19:08

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 maps cycles to boundaries. Thus it is zero on homology, (which is exactly the point of this construction), so $$f_*=g_*.$$

In your case, $g$ is a constant map. Your confusion comes, because you think that if $g$ is constant, then it is zero on cohomology. 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 $i>0$.

• The explicit use of $g$ and its effect on homology really made this click for me. – Henry Swanson Dec 2 '17 at 12:46