Fiber and limit isomorphisms force a base-axis isomorphism

Induct on $k$, assuming the base-axis maps are isomorphisms through column $k$. Column zero starts this induction. The tensor condition implies isomorphism on all $E_2^{p,q}$ with $p\le k$.

Induction on $r\ge 2$ gives simultaneously: $(A_r){\mathrm{\Phi }}_r^{p,q}\text{ is an isomorphism if }p\le k-r+1,(B_r){\mathrm{\Phi }}_r^{p,q}\text{ is injective if }p\le k\mathrm{.}$ For $r=2$ these follow from the preceding $E_2$ isomorphisms. Write $Z_r=\ker d_r$ and $B_r=\mathrm{im}{d}_r$, distinguishing the boundary group $B_r$ from the assertion $(B_r)$ by context. If $p\le k-r$, the domain map is an isomorphism and the outgoing-target map in column $p+r\le k$ is injective, so the map on cycles $Z_r^{p,q}$ is an isomorphism. If $p\le k-r+1$, both the domain and incoming-source maps are isomorphisms by $(A_r)$, so the maps on boundary groups and on the quotients by those boundaries are isomorphisms. Quotienting cycles by boundaries gives $(A_{r+1})$. For $p\le k$, the cycle map is injective by $(B_r)$, and the incoming-source map is surjective since $p-r\le k-r+1$. Thus every target boundary in the image of a source cycle lifts to a source boundary; the map on cycle/boundary quotients is injective. This gives $(B_{r+1})$.

Now consider the unknown bottom position $(k+1,0)$. It has no outgoing differential, and for each $r$ there is an exact sequence $Z_r^{k-r+1,r-1}⟶E_r^{k-r+1,r-1}\stackrel{d_r}{\to }{E}_r^{k+1,0}⟶E_{r+1}^{k+1,0}⟶0.$ The second term is mapped isomorphically by $(A_r)$. We claim the cycle term is mapped surjectively. Put $(u,v)=(k-r+1,r-1)$. For $s\ge r$, use $E_s^{u-s,v+s-1}⟶Z_s^{u,v}⟶E_{s+1}^{u,v}⟶0.$ Here the first map is the incoming differential restricted to cycles; this is exact because $d_s^2=0$. The map on the first term is an isomorphism by $(A_s)$, since $u-s=k-r-s+1\le k-s+1$. At all pages later than $s$, outgoing differentials from $(u,v)$ have negative fiber target, so $E_{s+1}^{u,v}=Z_{s+1}^{u,v}$. Start from stationarity, where its map is an isomorphism by the assumed limiting isomorphism, and descend on $s$; lifting an element of the third term and correcting by an element of the first proves surjectivity on $Z_s^{u,v}$, including $s=r$. The first quadrant gives a finite stationarity bound at every position, so this is finite downward induction.

Finally the bottom position itself is stationary for $r>k+1$. Starting from its limiting isomorphism, descend on $r$ in the displayed five-term exact sequence. Surjectivity on the first term and isomorphisms on the second and fourth show the third map is an isomorphism: for surjectivity lift its quotient in the fourth term and correct the difference by the second; for injectivity first lift a kernel element to the second, then lift its image in the first and subtract, using injectivity of the second. Thus ${\mathrm{\Phi }}_2^{k+1,0}$ is an isomorphism. Induction on $k$ proves the claim.