# First page of a spectral sequence

I have a very basic problem when trying to understand "You could have invented spectral sequences" by Timothy Chow (but I will index cohomologically since I'm more interested in cohomology). This is basic linear algebra and probably I miss something quite obvious.

We take a filtered complex a complex $\dots \to C^n \stackrel{d^n}{\to} C^{n+1} \to \dots$ of finite dimensional vector spaces, with a one-step filtration $F^1C \subset C$.

We define $E_0^{0,n} = C^n/F^1C^n$ and $E_0^{1,n -1} = F^1C^n$. We define $$E_1^{0,n} = \frac{\ker (E_0^{0,n} \to E_0^{0,n+1})}{\text{im}(E_0^{0,n-1} \to E_0^{0,n})}$$ where the maps are induced by $d$ and similarly $$E_1^{1,n} = \frac{\ker (E_0^{1,n} \to E_0^{1,n+1})}{\text{im}(E_0^{1,n-1} \to E_0^{1,n})}$$

Now I would like to check that $d$ induces map $E_1^{0,n} \to E_1^{1,n}$. I am not convinced by this. Indeed this map should be a map $$\frac{\ker (C^n/F^1C^n \to C^{n+1}/F^1C^{n+1})}{\text{im}(C^{n-1}/F^1C^{n-1} \to C^n/F^1C^n)} \to \frac{\ker (F^1C^{n+1} \to F^1C^{n+2})}{\text{im}(F^1C^n \to F^1C^{n+1})}$$

I don't see how to get an element of $F^1C^{n+1}$ from the kernel of $\overline d : C^n/F^1C^n \to C^{n+1}/F^1C^{n+1}$. If $\overline d$ lifts to a map $d : C^n/F^1C^n \to C^{n+1}$ then elements of $\overline d$ would be canonically send to some elements in $F^1C^{n+1}$. But I believe this is not the case, since there is no reasons that $F^1C^n$ goes to zero by $d$. So taking $dz$ just gives me an element of $C^3$. It would work if e.g we fix an arbitrary complement $K^n \oplus F^1C^n = C^n$ for each $n$ but we didn't do this, so I'm probably missing something really obvious or wrote something wrong.

Let $[x] \in \ker(C^n / F^1 C^n \to C^{n+1} / F^1 C^{n+1})$.
This means, by definition of the kernel, that you have a $x$ in $C^n$ such that $[dx] = 0$ in $C^{n+1}/F^1 C^{n+1}$. By definition of the quotient, this means that $dx$ is actually in $F^1 C^{n+1}$. So there's your element of $F^1 C^{n+1}$: it's just $dx$.
Next you need to check that $dx$ is actually in $\ker(d : F^1 C^{n+1} \to F^1 C^{n+2})$. But of course this follows from $d^2 = 0$. So from $x$ you do get an element of $\ker(F^1 C^{n+1} \to F^1 C^{n+2})$.
Moreover, suppose that $[x] = 0$, i.e. $x \in F^1 C^n$. You want to check that $dx = 0$. But in the target you mod out by the image $\operatorname{im}(F^1 C^n \to F^1 C^{n+1})$, so you're good.
At this point, we have a well defined linear map $$\ker(C^n / F^1 C^n \to C^{n+1} / F^1 C^{n+1}) \to \frac{\ker(F^1 C^{n+1} \to F^1 C^{n+2})}{\operatorname{im}(F^1 C^n \to F^1 C^{n+1})}.$$ The final step is to check that if $x$ is in the image of $C^{n-1}/F^1 C^{n-1} \to C^n / F^1 C^n$, then $[dx] = 0$. Again this follows from $d^2 = 0$ and we're done building the linear map.
• Ok, I forgot that moding out by $im(F^1C^n \to F^1C^{n+1})$ was making the map well-defined. Thank you ! – student Apr 3 '18 at 13:23