Spectral Sequences - John McCleary Example 1.A (First Quadrant Topological Spectral Sequence)

On John McCleary's book "A user's guide to sepctral sequences", on page 6 he gives the following example:

Example 1.A. Suppose that there is a first quadrant spectral sequence of cohomologica type with initial term $(E_2^{*,*},d_2)$, converging to the graded vector space $H^*$, and satisfying $F^{p+k}H^p=\{0\}$ for all $k>0$. Then $H^0=E_2^{0,0}$, and there is an exact sequence, $$0 \rightarrow E^{1,0}_2 \rightarrow H^1 \rightarrow E^{0,1}_2 \rightarrow E^{2,0}_2 \rightarrow E^{2,0}_\infty \rightarrow 0,$$ with $E^{2,0}_\infty$ a submodule of $H^2$.

He proceeds to prove this. First the proof that $H^0=E_2^{0,0}$ is really simple. Then he uses the filtration $$\{0\}\subset F^1H^1 \subset F^0H^1 = H^1\ \ \ \ \ \text{and} \ \ \ \ \ \{0\}\subset F^2H^2 \subset F^1H^2 \subset F^0H^2 = H^2$$ and says that this leads to the short exact sequences $$0 \rightarrow E_2^{1,0}\rightarrow H^1 \rightarrow E_2^{0,1} \rightarrow 0 \ \ \ \ \ \text{and}\ \ \ \ \ \ 0\rightarrow E^{2,0}_\infty \rightarrow H^2$$.

This is the part where I had a bit of problem with. The sequence on the right is quite simple, correct me if I'm wrong, but for that one you simply use the fact that the spectral sequence converges to $H^*$, thus $$E^{2,0}_\infty = \frac{F^2H^2}{F^3H^2} = \frac{F^2H^2}{0} = F^2H^2\subset H^2 $$ hence, $0\rightarrow E^{2,0}_\infty \hookrightarrow H^2$ is exact.

For the oder sequence I think we have to use both facts that $$E_2^{1,0}=E_\infty^{1,0}=\frac{F^1H^1}{F^2H^1} = \frac{F^1H^1}{0} = F^1H^1\subset H^1 $$ and $$\frac{F^0H^1}{F^1H^1} = \frac{H^1}{F^1H^1} = \frac{H^1}{E_2^{1,0}} = E^{0,1}_\infty = \ker(d_2:E^{0,1}_2\rightarrow E^{2,0}_2)$$ so we conlude that $$\frac{H^1}{E_2^{1,0}} = \ker(d_2)\subset E_2^{0,1}$$ therefore we have the exact sequence $$0 \rightarrow E_2^{1,0}\rightarrow H^1 \rightarrow E_2^{0,1} \rightarrow E_2^{2,0},$$ and, unless for some reason you can argue that $E_2^{2,0}=0$ you won't get the exact sequence mentioned in the book.

But this is acuatually no problem for getting the end result, since we already arrived almost at the desired exact sequence. All that is left to do is consider $$E^{2,0}_\infty = E^{2,0}_3 = \frac{\ker(d_2:E_2^{2,0}\rightarrow 0)}{\text{Im}(d_2:E^{0,1}_2\rightarrow E^{2,0}_2)} = \frac{E^{2,0}_2}{\text{Im}(d_2)}$$, hence we can define a map (a projection) $E^{2,0}_2 \rightarrow E^{2,0}_\infty$, with its kernel being exactly $\text{Im}(d_2)$, thus we get the exact sequence $$0 \rightarrow E_2^{1,0}\rightarrow H^1 \rightarrow E_2^{0,1} \rightarrow E_2^{2,0} \rightarrow E^{2,0}_\infty \rightarrow 0$$, which is exactly what we wanted.

So I have no problem in solving this example, I just wanted to confirm if that sequence $0 \rightarrow E_2^{1,0}\rightarrow H^1 \rightarrow E_2^{0,1} \rightarrow 0$ is actually not exact. Maybe I made some mistake in my solution and somehow got to the correct solution anyway. But I really don't see how that sequence could be exact without some other assumption about the spectral sequence.

Solution 1:

That’s right: since $E_2^{0,1}$ isn’t stable without extra assumptions on $d_2: E_2^{0,1} \to E_2^{2,0}$, McCleary probably meant this short exact sequence to be $0\to E_2^{1,0}\to H^1 \to E_3^{0,1} \to 0$, using $E_3^{0,1} = E_\infty^{0,1}$, and then concluding via $E_3^{0,1} =\operatorname{ker}(d_2: E_2^{0,1} \to E_2^{2,0})$ exactly as you describe.