Bourbaki

Theorem 6

Tag 05SG, permanent. Cite it as [Bourbaki, Tag 05SG].

Algebra, chapter II, § 7. Vector spaces

For every vector space E over a field K, the canonical mapping cE:E→E∗∗c_E : E \to E^{**} (§ 2, no. 7) is injective; for it to be bijective, it is necessary and sufficient that E be finite-dimensional.

The first assertion and the fact that if E is finite-dimensional cEc_E is bijective are special cases of § 2, no. 7, Proposition 14. Suppose that E is infinite-dimensional, so that we may assume that E=Ks(L)E = K_s^{(L)}, where L is an infinite set and therefore E∗=KdLE^* = K_d^L. Let (eλ)λ∈L(e_\lambda)_{\lambda \in L} be the canonical basis of E and let (eλ∗)λ∈L(e_\lambda^*)_{\lambda \in L} be the corresponding family of coordinate forms in E∗E^* (§ 2, no. 6); the vector subspace of E∗E^* generated by the eλ∗e_\lambda^* is just the direct sum F′=Kd(L)F' = K_d^{(L)} and the hypothesis that L is infinite implies that F′≠E∗F' \neq E^*. Then there exists a hyperplane H′H' of E∗E^* containing F′F' (no. 3, Proposition 8) and, as E∗/H′E^*/H' is non-zero, its dual is also non-zero (Corollary to Theorem 4), which is identified with the orthogonal H′′H'' of H′H' in E∗∗E^{**} (§ 2, no. 6, Corollary to Proposition 9). But H′′∩cE(E)H'' \cap c_E(E) is contained in the image under cEc_E of the orthogonal of F′F' in E, which is by definition 0; hence cE(E)=E∗∗c_E(E) = E^{**} is impossible.

E will usually be identified with the subspace of E∗∗E^{**} the image of cEc_E.

Let E, F be two vector spaces over a field K and u:E→Fu : E \to F a linear mapping. We shall define canonical isomorphisms:

(1) Of the dual of Im(u) = u(E) onto Im(tu^t u) = tu(F∗)^t u(F^*).

(2) Of the dual of Ker(u) = −1u(0)^{-1}u(0) onto Coker(tu^t u) = E∗/tu(F∗)E^*/^t u(F^*).

(3) Of the dual of Coker(u) = F/u(E) onto Ker(tu^t u) = tu−1(0)^t u^{-1}(0).

We write I = Im(u), N = Ker(u), C = Coker(u); from the exact sequences

(14) 0→N→E→pI→0,0→I→jF→C→00 \to N \to E \xrightarrow{p} I \to 0,\quad 0 \to I \xrightarrow{j} F \to C \to 0

we derive, by transposition (Corollary to Theorem 5), the exact sequences

(15)

0→I∗→tpE∗→N∗→0,0→C∗→F∗→tjI∗→0. \begin{array}{ccccccccc} 0 & \to & I^* & \xrightarrow{^t p} & E^* & \to & N^* & \to & 0, \\ 0 & \to & C^* & \to & F^* & \xrightarrow{^t j} & I^* & \to & 0. \end{array}

Moreover, as u=j∘p, tu=tp∘tju = j \circ p,\ ^t u = ^t p \circ ^t j; the exact sequences (15) thus define canonical isomorphisms of C∗C^* onto Ker(tu^t u), of I∗I^* onto Im(tu^t u) and of N∗N^* onto Coker(tu^t u), since tp^t p is injective and tj^t j surjective. To be precise, let y∈Im(u)y \in \mathrm{Im}(u), z∈Ker(u)z \in \mathrm{Ker}(u), t∈Coker(u)t \in \mathrm{Coker}(u), y′⊂Im(tu)y' \subset \mathrm{Im}(^t u), z′∈Coker(tu)z' \in \mathrm{Coker}(^t u), t′∈Ker(tu)t' \in \mathrm{Ker}(^t u); when y′y', z′z', t′t' are canonically identified with linear forms on Im(u)\mathrm{Im}(u), Ker(u)\mathrm{Ker}(u) and Coker(u)\mathrm{Coker}(u) respectively, then

(16) ⟨y,y′⟩=⟨x,y′⟩\langle y, y' \rangle = \langle x, y' \rangle for all x∈Ex \in E such that u(x)=yu(x) = y;

(17) ⟨z,z′⟩=⟨z,x∗⟩\langle z, z' \rangle = \langle z, x^* \rangle for all x∗∈E∗x^* \in E^* whose class mod. tu(F)^t u(F) is equal to z′z';

(18) ⟨t,t′⟩=⟨s,t′⟩\langle t, t' \rangle = \langle s, t' \rangle for all s∈Fs \in F whose class mod. u(E)u(E) is equal to tt.

In particular we derive from these results:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Transcribed from the printed edition, 1998, Springer. Not translated.