# spectral sequence example diagonal map confusion

I'm attempting to wrap my head around spectral sequences, so constructed a really basic example to apply the definitions and go through the motions. My filtered chain complexes are:

$$F_2C_*: 0 \rightarrow \mathbb{Z} \xrightarrow{i_1} \mathbb{Z_1}\oplus \mathbb{Z_2} \xrightarrow{p_2} \mathbb{Z} \rightarrow 0$$

$$F_1C_*: 0 \rightarrow 0 \rightarrow \mathbb{Z}_2 \xrightarrow{p_2} \mathbb{Z} \rightarrow 0$$

$$F_0C_*:$$ The zero complex.

Here $$i_1$$ and $$p_2$$ are the inclusion and projection maps, and the sub-indices on the $$\mathbb{Z}s$$ are just for labeling purposes.

With $$Z^r_{p,q}$$ defined as $$\{x \in F_pC_q : \partial x \in F_{p-r}C_{q-1}\}/F_{p-1}C_q$$, the differential is meant to restrict to well-defined maps $$d^r:Z^r_{p,q} \rightarrow Z^r_{p-r,q-1}$$.

But in my example the map $$Z^1_{2,2} \rightarrow Z^1_{1,1}$$ does not appear to be well-defined:

$$Z^1_{2,2} := \{x \in \mathbb{Z}\oplus \mathbb{Z} : p_2 x \in \mathbb{Z} \}/\mathbb{Z}_2 \cong \mathbb{Z}_1$$,

$$Z^1_{1,1} := \mathbb{Z}$$.

However $$[(a,b)]$$ and $$[(a,b')]$$ in $$Z^1_{2,2}$$ get sent to $$b$$ and $$b'$$ in $$\mathbb{Z}$$ respectively, even though $$[(a,b)] = [(a,b')]$$.

I cannot see where I've made a mistake, but I'm pretty sure something's not right. Where have I gone wrong?

Given a map $$A \xrightarrow{f} C$$ with $$B \subset A$$ and $$D \subset C$$, in order for the quotient map $$A/B \xrightarrow{\bar f} C/D$$ to be defined we must have $$f(B) \subset D$$.

But it this case we have a map $$(Z_1 \oplus Z_2) / Z_2 \xrightarrow{\bar p_2} Z/\{0\}$$, but $$p_2(Z_2) = Z_2$$ which is not contained in $$\{0\}!$$

## 1 Answer

$$Z_{p,q}^r$$ isn’t a subquotient just a subobject. $$Z_{p,q}^r := \{x \in F^pC_q: dx \in F^{p-r}C_{q-1}\}$$ check eg wikipedia.