Sentence examples for natural maps from inspiring English sources

Exact(6)

The invariant consists ofK0,K1, theK0-group with Z/pcoefficients, the order structures these groups possess, and the natural maps between the three groups.

It's immediate from the definitions that an elementary extension of an elementary extension of A is again an elementary extension of A. Elementary embeddings are natural maps to consider within first-order model theory.

Note that since H e ´ t 1 ( R n, S ) = 1, the natural maps N H ( L ) ( R n ) → H ′ ( R n ) and N H ( L ) ( F n ) → H ′ ( F n ) are surjective.

In the general case, by induction, we have a series of objects (gamma _* tau _{le n}^*E)) related by natural maps begin{aligned} tau _{le n-1}^*gamma _*left( tau _{le n}^*Eright) rightarrow gamma _*left( tau _{le n-1}^*Eright), end{aligned} (8.19 and by adjunction, these maps are isomorphisms.

Therefore by adjunction, we have natural maps begin{aligned} overline{L}_i circ R^cdot Phi rightarrow R^cdot Phi (i) circ overline{L}_i, qquad R^cdot Phi (i) circ overline{M}_i rightarrow overline{M}_i circ R^cdot Phi, end{aligned} (8.38 and by (8.18) and (8.25), these maps are isomorphisms.

6.16] S R p and S R p ′ are conjugate under G R p ( R p ) = G ( R p ), Thus the image of α under the composition of the natural maps H e ´ t 1 ( R, Z G ( S ) ) → H e ´ t 1 ( R, N G ( S ) ) → H e ´ t 1 ( R p, N G R p ( S R p ) ) → H e ´ t 1 ( R p, G R p ) is trivial.

Similar(54)

People not blessed with natural mapping ability can compensate by using the sort of "metacognitive strategies" taught in Boy Scouts, like placing a pile of rocks by a path one has already taken, assuming one can find rocks in a corn field.

People not blessed with natural mapping ability can _________ by using the sort of "metacognitive strategies" taught in Boy Scouts, like placing a pile of rocks by a path one has already taken, assuming one can find rocks in a corn field.

Note that the natural map induces (2.29).

Hence, we have a natural map (3.2).

The natural map m ⊗ k R → R m is an R –module isomorphism.

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