By Mao L.

ISBN-10: 1599731002

ISBN-13: 9781599731001

This monograph is stimulated with surveying arithmetic and physics via CC conjecture, i.e., a mathematical technological know-how should be reconstructed from or made through combinatorialization. subject matters lined during this publication comprise primary of mathematical combinatorics, differential Smarandache n-manifolds, combinatorial or differentiable manifolds and submanifolds, Lie multi-groups, combinatorial relevant fiber bundles, gravitational box, quantum fields with their combinatorial generalization, additionally with discussions on basic questions in epistemology. All of those are priceless for researchers in combinatorics, topology, differential geometry, gravitational or quantum fields.

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4. u1 u4 G u2 u1 u2 u1 u3 u3 u4 u3 G1 G2 u2 u4 G3 Fig. 4 For a nonempty subset U of the vertex set V (G) of a graph G, the subgraph U of G induced by U is a graph having vertex set U and whose edge set consists of these edges of G incident with elements of U. 4 Graphs induced if H ∼ = U for some subset U of V (G). Similarly, for a nonempty subset F of E(G), the subgraph F induced by F in G is a graph having edge set F and whose vertex set consists of vertices of G incident with at least one edge of F .

Proof If |X| = |Y |, there are no bijections from X to Y by deﬁnition. Whence, we only need to consider the case of |X| = |Y |. Let X = {x1 , x2 , · · · , xn } and Y = {y1 , y2 , · · · , yn }. For any permutation p on y1 , y2 , · · · , yn , the mapping determined by x1 x2 ··· xn p(y1) p(y2 ) · · · p(yn ) is a bijection from X to Y , and vice versa. Whence, |Bij(Y X )| = 0 if |X| = |Y |, n! = |Y |! if |X| = |Y | Similarly, if |X| > |Y |, there are no injections from X to Y by deﬁnition. Whence, we only need to consider the case of |X| ≤ |Y |.

It can be veriﬁed immediately that (h ◦ f )−1 = f −1 ◦ h−1 by deﬁnition. We have a characteristic for bijections from X to Y by composition operations. 1 A mapping f : X → Y is a bijection if and only if there exists a mapping h : Y → X such that f ◦ h = 1Y and h ◦ f = 1X . Proof If f is a bijection, then for ∀y ∈ Y , there is a unique x ∈ X such that f (x) = y. Deﬁne a mapping h : Y → X by h(y) = x for ∀y ∈ Y and its correspondent x. Then it can be veriﬁed immediately that f ◦ h = 1Y and h ◦ f = 1X .

### Combinatorial geometry with application to field theory by Mao L.

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