Functor - Revision history

Vipul at 23:22, 9 December 2008

2008-12-09T23:22:43Z

← Older revision		Revision as of 23:22, 9 December 2008
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	Suppose <math>\mathcal{C},\mathcal{D}</math> are [[category\|categories]]. A '''functor''' (also called '''covariant functor''') <math>\mathcal{F}</math> from <math>\mathcal{C}</math> to <math>\mathcal{D}</math> comprises the following data:		Suppose <math>\mathcal{C},\mathcal{D}</math> are [[category\|categories]]. A '''functor''' (also called '''covariant functor''') <math>\mathcal{F}</math> from <math>\mathcal{C}</math> to <math>\mathcal{D}</math> comprises the following data:

	* A mapping <math>\mathcal{F}: \operatorname{Ob}\mathcal{C} \to \operatorname{Ob}\mathcal{D}</math>.		* '''Object level mapping''': A mapping <math>\mathcal{F}: \operatorname{Ob}\mathcal{C} \to \operatorname{Ob}\mathcal{D}</math>.
	* For any <math>A,B \in \mathcal{C}</math>, a mapping <math>\mathcal{F}: \mathcal{C}(A,B) \to \mathcal{D}(FA,FB)</math>.		* '''Morphism level mapping''': For any <math>A,B \in \mathcal{C}</math>, a mapping <math>\mathcal{F}: \mathcal{C}(A,B) \to \mathcal{D}(FA,FB)</math>.

	satisfying the following condition:		satisfying the following condition:

	* For any <math>A \in \mathcal{C}</math>, <math>\mathcal{F}(\operatorname{id}_A) = \operatorname{id}_{F(A)}</math>.		* '''It preserves the identity morphism''': For any <math>A \in \mathcal{C}</math>, <math>\mathcal{F}(\operatorname{id}_A) = \operatorname{id}_{F(A)}</math>.
	* For any <math>A,B,C \in \mathcal{C}</math>, and <math>f \in \mathcal{C}(A,B), g \in \mathcal{C}(B,C)</math>, we have <math>F(g \circ f) = Fg \circ Ff</math>.		* '''It preserves composition''': For any <math>A,B,C \in \mathcal{C}</math>, and <math>f \in \mathcal{C}(A,B), g \in \mathcal{C}(B,C)</math>, we have <math>F(g \circ f) = Fg \circ Ff</math>.

	There is a related notion of [[contravariant functor]]. Note that a contravariant functor between two categories is ''not'' a functor between the categories. To emphasize that a functor is a functor and not a contravariant functor, we use the term ''covariant functor''.		There is a related notion of [[contravariant functor]]. Note that a contravariant functor between two categories is ''not'' a functor between the categories. To emphasize that a functor is a functor and not a contravariant functor, we use the term ''covariant functor''.

Vipul at 23:19, 9 December 2008

2008-12-09T23:19:02Z

← Older revision		Revision as of 23:19, 9 December 2008
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			{{basicdef}}
	==Definition==		==Definition==

Vipul at 23:18, 9 December 2008

2008-12-09T23:18:48Z

← Older revision		Revision as of 23:18, 9 December 2008
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	==Definition==		==Definition==

	Suppose <math>\mathcal{C},\mathcal{D}</math> are [[category\|categories]]. A '''functor''' <math>\mathcal{F}</math> from <math>\mathcal{C}</math> to <math>\mathcal{D}</math> comprises the following data:		Suppose <math>\mathcal{C},\mathcal{D}</math> are [[category\|categories]]. A '''functor''' (also called '''covariant functor''') <math>\mathcal{F}</math> from <math>\mathcal{C}</math> to <math>\mathcal{D}</math> comprises the following data:

	* A mapping <math>\mathcal{F}: \operatorname{Ob}\mathcal{C} \to \operatorname{Ob}\mathcal{D}</math>.		* A mapping <math>\mathcal{F}: \operatorname{Ob}\mathcal{C} \to \operatorname{Ob}\mathcal{D}</math>.
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	* For any <math>A \in \mathcal{C}</math>, <math>\mathcal{F}(\operatorname{id}_A) = \operatorname{id}_{F(A)}</math>.		* For any <math>A \in \mathcal{C}</math>, <math>\mathcal{F}(\operatorname{id}_A) = \operatorname{id}_{F(A)}</math>.
	* For any <math>A,B,C \in \mathcal{C}</math>, and <math>f \in \mathcal{C}(A,B), g \in \mathcal{C}(B,C)</math>, we have <math>F(g \circ f) = Fg \circ Ff</math>.		* For any <math>A,B,C \in \mathcal{C}</math>, and <math>f \in \mathcal{C}(A,B), g \in \mathcal{C}(B,C)</math>, we have <math>F(g \circ f) = Fg \circ Ff</math>.

			There is a related notion of [[contravariant functor]]. Note that a contravariant functor between two categories is ''not'' a functor between the categories. To emphasize that a functor is a functor and not a contravariant functor, we use the term ''covariant functor''.

Vipul: New page: ==Definition== Suppose $\mathcal{C},\mathcal{D}$ are categories. A '''functor''' $\mathcal{F}$ from $\mathcal{C}$ to $\mathcal{D}$ ...

2008-12-09T00:12:11Z

New page: ==Definition== Suppose <math>\mathcal{C},\mathcal{D}</math> are categories. A '''functor''' <math>\mathcal{F}</math> from <math>\mathcal{C}</math> to <math>\mathcal{D}</math>...

New page

==Definition==

Suppose <math>\mathcal{C},\mathcal{D}</math> are [[category|categories]]. A '''functor''' <math>\mathcal{F}</math> from <math>\mathcal{C}</math> to <math>\mathcal{D}</math> comprises the following data:

* A mapping <math>\mathcal{F}: \operatorname{Ob}\mathcal{C} \to \operatorname{Ob}\mathcal{D}</math>.
* For any <math>A,B \in \mathcal{C}</math>, a mapping <math>\mathcal{F}: \mathcal{C}(A,B) \to \mathcal{D}(FA,FB)</math>.

satisfying the following condition:

* For any <math>A \in \mathcal{C}</math>, <math>\mathcal{F}(\operatorname{id}_A) = \operatorname{id}_{F(A)}</math>.
* For any <math>A,B,C \in \mathcal{C}</math>, and <math>f \in \mathcal{C}(A,B), g \in \mathcal{C}(B,C)</math>, we have <math>F(g \circ f) = Fg \circ Ff</math>.

Functor - Revision history

Vipul at 23:22, 9 December 2008

Vipul at 23:19, 9 December 2008

Vipul at 23:18, 9 December 2008

Vipul: New page: ==Definition== Suppose \mathcal{C},\mathcal{D} are categories. A '''functor''' \mathcal{F} from \mathcal{C} to \mathcal{D}...

Vipul: New page: ==Definition== Suppose $\mathcal{C},\mathcal{D}$ are categories. A '''functor''' $\mathcal{F}$ from $\mathcal{C}$ to $\mathcal{D}$ ...