Remark 13.9.7. To see the last displayed equality in the proof above we can argue with elements as follows. We have $s\pi (l, k, k^{+}) = (l, 0, 0)$. Hence the morphism of the left hand side maps $(l, k, k^{+})$ to $(0, k, k^{+})$. On the other hand $h(l, k, k^{+}) = (0, 0, k)$ and $d(l, k, k^{+}) = (dl, dk + k^{+}, -dk^{+})$. Hence $(dh + hd)(l, k, k^{+}) = d(0, 0, k) + h(dl, dk + k^{+}, -dk^{+}) = (0, k, -dk) + (0, 0, dk + k^{+}) = (0, k, k^{+})$ as desired.

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