Edge maps of a bounded skeletal AHSS

For cohomology use homological indices $(a,b)$, putting $D_{a,b}=h^{-a-b-1}(X^{-a-1})$ and $E_{a,b}=h^{-a-b}(X^{-a},X^{-a-1})$. The maps $i,j,k$ are respectively restriction, pair boundary and pair-to-absolute map. The given pair sequences prove all three exactness conditions of [F1]. For homology put $D_{p,q}=h_{p+q}(X^p)$ and $E_{p,q}=h_{p+q}(X^p,X^{p-1})$; inclusion, pair map and boundary are $i,j,k$, and again the pair sequences prove exactness. Thus [F2] applies to both couples. In the cohomological output reindex $(a,b)=(-p,-q)$.

Fix $0\le p\le N$, $q\in \mathbb{Z}$ and $n=p+q$. Substituting the cohomological groups of step 1.1 in [F2] gives $N^r=k^{-1}(\mathrm{im}(h^n(X^{p+r-1})\to h^n(X^p)))$ and $B^r=j(\ker (h^{n-1}(X^{p-1})\to h^{n-1}(X^{p-r})))$. For $r\ge N+2$, the former skeleton is $X$ and the latter is empty. Hence $N^{\mathrm{\infty }}=k^{-1}(\mathrm{im}(h^n(X)\to h^n(X^p)))$ and $B^{\mathrm{\infty }}=\mathrm{im}j=\ker k$. Put $K=\mathrm{im}(h^n(X)\to h^n(X^p))\cap \ker (h^n(X^p)\to h^n(X^{p-1}))$. Exactness says $\mathrm{im}k$ is the second factor, so $k$ maps $N^{\mathrm{\infty }}$ onto $K$ with kernel $B^{\mathrm{\infty }}$. Restriction maps $F^p{h}^n(X)$ onto $K$: a lift $x$ of a member of $K$ restricts to zero on $X^{p-1}$. Its kernel is $F^{p+1}$. The two quotient isomorphisms give exactly the claimed map, with its lift independent modulo $F^{p+1}$.

In homology [F2] gives $N^r=k^{-1}(\mathrm{im}(h_{n-1}(X^{p-r})\to h_{n-1}(X^{p-1})))$ and $B^r=j(\ker (h_n(X^p)\to h_n(X^{p+r-1})))$. Thus for $r\ge N+2$, $N^{\mathrm{\infty }}=\ker k=\mathrm{im}j$ and $B^{\mathrm{\infty }}=j(\ker u)$, where $u:h_n(X^p)\to h_n(X)$. Define $\mathrm{im}j\to F_p\mathrm{/}{F}_{p-1}$ by $j(x)\mapsto u(x)+F_{p-1}$. Two lifts differ by $\ker j=\mathrm{im}(h_n(X^{p-1})\to h_n(X^p))$, so this is well defined and surjective. Its kernel equals $j(\ker u)$: if $u(x)$ comes from $y\in h_n(X^{p-1})$, subtract the image of $y$ from $x$ to get a lift with zero image in $h_n(X)$ and unchanged $j(x)$. Conversely such a lift plainly maps to zero. Quotienting gives $E_{p,q}^{\mathrm{\infty }}\cong F_p\mathrm{/}{F}_{p-1}$. Its inverse is induced by $j$, not by a map from $E^1$ directly to $h_n(X)$.

The cohomological filtration has $F^0=h^n(X)$ and $F^{N+1}=0$, since restriction to the empty space is zero and restriction to $X$ is the identity. At $p=0$, step 2.1 identifies the quotient by $F^1$ with the image of restriction to $X^0$ and hence with $E_{\mathrm{\infty }}^{0,n}$. At $p=N$, its lift $x$ is already a class on $X$, yielding the inclusion $E_{\mathrm{\infty }}^{N,n-N}\cong F^N\hookrightarrow h^n(X)$. These are precisely the stated cohomological edges.

Likewise $F_{-1}=0$ and $F_N=h_n(X)$. At $p=0$ step 2.2 gives $E_{0,n}^{\mathrm{\infty }}\cong F_0$ followed by the inclusion in $h_n(X)$; the composite from $h_n(X^0)$ is the original skeletal inclusion map. At $p=N$, the map induced by $j$ is the quotient $h_n(X)\to h_n(X)\mathrm{/}{F}_{N-1}\cong E_{N,n-N}^{\mathrm{\infty }}$. Thus the homological edges have the asserted directions.

The same bound $r\ge N+2$ worked for every $q$ in steps 2.1 and 2.2. Outside $0\le p\le N$ the pair terms vanish by exactness for an identical pair, so all their later subquotients vanish. No boundedness on $q$ is used. If $N=0$, the two edges coincide with the identity under the displayed identifications. If $X$ is empty, exactness and the zero absolute groups make all relative groups and pages zero. The representative arguments establish unique cosets and never choose a family of lifts; no additional choice principle is used.