MyWikiBiz, Author Your Legacy — Tuesday May 07, 2024
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, 11:40, 27 March 2009
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| {| align="center" cellpadding="8" width="90%" <!--QUOTE--> | | {| align="center" cellpadding="8" width="90%" <!--QUOTE--> |
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− | <p>'''Definition 1.3.''' A ''functor'' <math>F : \mathcal{A} \to \mathcal{B}</math> is first of all a morphism of graphs (see Example C4), that is, it sends objects of <math>\mathcal{A}</math> to objects of <math>\mathcal{B}</math> and arrows of <math>\mathcal{A}</math> to arrows of <math>\mathcal{B}</math> such that, if <math>f : A \to A',</math> then <math>F(f) : F(A) \to F(A').</math> Moreover, a functor preserves identities and composition; thus:</p> | + | <p>'''Definition 1.3.''' A ''functor'' <math>F : \mathcal{A} \to \mathcal{B}</math> is first of all a morphism of graphs …, that is, it sends objects of <math>\mathcal{A}</math> to objects of <math>\mathcal{B}</math> and arrows of <math>\mathcal{A}</math> to arrows of <math>\mathcal{B}</math> such that, if <math>f : A \to A',</math> then <math>F(f) : F(A) \to F(A').</math> Moreover, a functor preserves identities and composition; thus:</p> |
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− | <p>F(1_A) = 1_F(A),</p> | + | ::<p><math>F(1_A) = 1_{F(A)}, \quad F(gf) = F(g)F(f).</math></p> |
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− | <p>F(gf) = F(g)F(f).</p> | + | <p>In particular, the identity functor <math>1_\mathcal{A} : \mathcal{A} \to \mathcal{A}</math> leaves objects and arrows unchanged and the composition of functors <math>F : \mathcal{A} \to \mathcal{B}</math> and <math>G : \mathcal{B} \to \mathcal{C}</math> is given by:</p> |
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− | <p>In particular, the identity functor 1_$A$ : $A$ -> $A$ leaves objects and arrows unchanged and the composition of functors F : $A$ -> $B$ and G : $B$ -> $C$ is given by:</p> | + | ::<p><math>(GF)(A) = G(F(A)), \quad (GF)(f) = G(F(f)),</math></p> |
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− | <p>(GF)(A) = G(F(A)),</p> | + | <p>for all objects <math>A\!</math> of <math>\mathcal{A}</math> and all arrows <math>f : A \to A'</math> in <math>\mathcal{A}.</math></p> |
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− | <p>(GF)(f) = G(F(f)),</p> | |
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− | <p>for all objects A of $A$ and all arrows f : A -> A' in $A$.</p>
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| <p>(Lambek & Scott, 6).</p> | | <p>(Lambek & Scott, 6).</p> |