# Truncating columns of a double complex to get a filtration of the homology of the total complex

My question refers to the document found here. Specifically page 394 of the book (page 14 of the pdf). Theorem 2.5 on that page refers to "the filtration of $H_{m}(Tot)$ obtained by truncating columns of the double complex".

I have no idea what this phrase is supposed to mean. What exactly does it mean by truncating the columns of the double complex? And how does this result in a filtration of the homology of the total complex?

So if $C$ is a double complex, we have by definition $Tot(C)_i = \bigoplus_{p+q = i} C_{p,q}$.
Now, we can look at the subcomplexe $Tot(C)^{p \geq r}$, verifying $Tot(C)_i = \bigoplus_{p+q = i, p \geq r} C_{p,q}$. If you draw $Tot(C)$ in the $xy$ plane, $Tot(C)^{p \geq r}$ is the subcomplex with zero at every $(p,q)$ with $p < r$, hence the name truncating the columns.
Notice that for each $r$ there are maps $Tot(C)^{p \geq r} \to Tot(C)$. This gives submodules $H_m^{p \leq r}(Tot(C)) \subset H_m(Tot(C))$ and this is how your filtration is defined.
• So the maps $Tot(C)^{p \geq r} \rightarrow Tot(C)$ are just the inclusion maps, right? But a monomorphism of short exact sequences surely doesn't result in a monomorphism on homology. Just consider the long exact sequence resulting from the short exact sequence of chain complexes. We would need the connecting map to be the zero map for that to be true. – Luke Aug 8 '18 at 1:31