Manuscript XVI-O1-ST@5 · XVI-O1-ST-v5.tex · section · dd45e0d1

Free exactness, tangent failure, and closed nonexact vectors

Retained exact synthetic examples › Free exactness, tangent failure, and closed nonexact vectors · lines 1885–1915

Let

S = 101011, B_{0} = (- 1, - 1, 1), A = (100100) .

Then $B_{0} S = 0$ , $A S = I_{2}$ , and

c (ξ) = ξ (1, - 1, 0)^{T}, d (ξ) = (- ξ, ξ)^{T}

satisfy cancellation and $c^{'} = S (1, - 1)^{T}$ . The vector $(0, 0, 1)^{T}$ is not exact.

With

A_{*} = 101000011, A_{*} S = 112011,

the left kernel is $(- 1, - 1, 1)$ . Target $(1, 0, 1)^{T}$ is feasible; $(1, 0, 0)^{T}$ is not. Also

S_{0} = e_{1}, B_{1} = (0, 0, 1), z = e_{2}

give $B_{1} S_{0} = B_{1} z = 0$ but $z \in / im S_{0}$ .

Displayed formulas in this passage