If 𝒜, ℬ, and 𝒞 are categories, then a bifunctor T: 𝒜 ×ℬ →𝒞 assigns, to each ordered pair of obje

If 𝒜, ℬ, and 𝒞 are categories, then a bifunctor T: 𝒜 ×ℬ →𝒞...

If $\mathcal{A}, \mathcal{B}$, and $\mathcal{C}$ are categories, then a bifunctor $T: \mathcal{A} \times \mathcal{B} \rightarrow \mathcal{C}$ assigns, to each ordered pair of objects $(A, B)$, where $A \in \mathrm{ob}(\mathcal{A})$ and $B \in o b(\mathcal{B})$, an object $T(A, B) \in o b(\mathcal{C})$, and to each ordered pair of morphisms $f: A \rightarrow A^{\prime}$ in $\mathcal{A}$ and $g: B \rightarrow B^{\prime}$ in $\mathcal{B}$, a morphism $T(f, g): T(A, B) \rightarrow T\left(A^{\prime}, B^{\prime}\right)$, such that (a) fixing either variable is a functor; for example, if $A \in \mathrm{ob}(\mathcal{A})$, then $T_{A}=T(A, \square): \mathcal{B} \rightarrow \mathcal{C}$ is a functor, where $T_{A}(B)=T(A, B)$ and $T_{A}(g)=T\left(1_{A}, g\right)$, (b) the following diagram commutes: $$ \begin{gathered} T(A, B) \stackrel{T\left(1_{A}, g\right)}{\longrightarrow} T\left(A, B^{\prime}\right) \\ T\left(f, 1_{B}\right) \mid \\ T\left(A^{\prime}, B\right) \underset{T\left(1_{A^{\prime}}, g\right)}{\longrightarrow} T\left(A^{\prime}, B^{\prime}\right) . \end{gathered} $$ (i) Prove that $\otimes: \operatorname{Mod}_{R} \times{ }_{R} \operatorname{Mod} \rightarrow \mathbf{A b}$ is a bifunctor. (ii) Prove that Hom: ${ }_{R}$ Mod $\times{ }_{R}$ Mod $\rightarrow \mathbf{A b}$ is a bifunctor if we modify the definition of bifunctor to allow contravariance in one variable.

If $\mathcal{A}, \mathcal{B}$, and $\mathcal{C}$ are categories, then a bifunctor $T: \mathcal{A} \times \mathcal{B} \rightarrow \mathcal{C}$ assigns, to each ordered pair of objects $(A, B)$, where $A \in \mathrm{ob}(\mathcal{A})$ and $B \in o b(\mathcal{B})$, an object $T(A, B) \in o b(\mathcal{C})$, and to each ordered pair of morphisms $f: A \rightarrow A^{\prime}$ in $\mathcal{A}$ and $g: B \rightarrow B^{\prime}$ in $\mathcal{B}$, a morphism $T(f, g): T(A, B) \rightarrow T\left(A^{\prime}, B^{\prime}\right)$, such that
(a) fixing either variable is a functor; for example, if $A \in \mathrm{ob}(\mathcal{A})$, then $T_{A}=T(A, \square): \mathcal{B} \rightarrow \mathcal{C}$ is a functor, where $T_{A}(B)=T(A, B)$ and $T_{A}(g)=T\left(1_{A}, g\right)$,
(b) the following diagram commutes:
$$
\begin{gathered}
T(A, B) \stackrel{T\left(1_{A}, g\right)}{\longrightarrow} T\left(A, B^{\prime}\right) \\
T\left(f, 1_{B}\right) \mid \\
T\left(A^{\prime}, B\right) \underset{T\left(1_{A^{\prime}}, g\right)}{\longrightarrow} T\left(A^{\prime}, B^{\prime}\right) .
\end{gathered}
$$
(i) Prove that $\otimes: \operatorname{Mod}_{R} \times{ }_{R} \operatorname{Mod} \rightarrow \mathbf{A b}$ is a bifunctor.
(ii) Prove that Hom: ${ }_{R}$ Mod $\times{ }_{R}$ Mod $\rightarrow \mathbf{A b}$ is a bifunctor if we modify the definition of bifunctor to allow contravariance in one variable.

Key Concepts

Hom Functor

The Hom functor assigns, to each pair of objects in a module category, the set (or abelian group) of morphisms between those objects. When considered as a bifunctor, it is typically contravariant in one variable and covariant in the other. The Hom functor is central to many areas of algebra and category theory because it encapsulates the notion of morphisms and provides a way to study the structure and relationships of modules in an abstract setting.

Contravariant Functor

A contravariant functor is similar to a functor, but it reverses the direction of morphisms. In other words, if a contravariant functor maps a morphism from object A to object B in the source category, it assigns a morphism from the image of B to the image of A in the target category. This reversal is an important concept, particularly in cases like the Hom functor, where one of its arguments may require contravariance to maintain the desired structure.

Tensor Product

The tensor product is an operation on modules (or vector spaces) that produces a new module (or vector space) from two given ones. It plays a key role in multilinear algebra and category theory, as it can be viewed as a bifunctor from the product of module categories to the category of abelian groups. Its universal property characterizes bilinear maps, making it a central construction in homological algebra and related fields.

Functor

A functor is a mapping between categories that preserves the categorical structure. This means it assigns to each object in the source category an object in the target category, and to each morphism in the source a corresponding morphism in the target, in such a way that the identities and the composition of morphisms are preserved. Functors are fundamental in relating different categories and transferring structures between them.

Category

A category is a structure consisting of a collection of objects and a collection of morphisms (or arrows) between these objects, together with rules for composing these morphisms that include associativity and the existence of identity morphisms for each object. Categories provide an abstract framework to study mathematical structures and their relationships in a unified way.

Bifunctor

A bifunctor is a generalization of a functor that operates on a product of two categories. It assigns an object in a target category to every ordered pair of objects from the two source categories and similarly assigns a morphism to every ordered pair of morphisms, all the while preserving the functorial structure in each variable. Bifunctors must also satisfy coherence conditions such as the commutativity of diagrams when morphisms are applied in different orders.

Transcript

00:01 Hello there.

00:02 Okay, so here we got two mappings, the map a from a to b, the map g from b to c, and the composition that should access this composition.

00:14 Okay, so we need to prove some statements.

00:18 The first statement say that if f and g are injective or one -to -one, then the composition is also injective.

00:39 Okay, so basically what we need to prove is that g off of x is equal to gof of y if and only if x is equal to y.

00:54 This will prove that the composition is injected.

00:59 So this is true or is false? okay, so what we know is that gof of x is equal to say g applied to the function.

01:16 F of x, right? it's just notation.

01:23 So what we know is that g is injective.

01:26 So g of f of x and g of f of y, because this part is because g is injective, this is equivalent to say that f of x is equals to f of y.

01:51 But also f is injective.

01:57 So this is equivalent to say that x is equals to y.

02:04 And we have shown that if we take g of g of f of x and g of f of y, these two are equal if and only if x is equal to y.

02:23 Exactly what we need to prove here.

02:26 So we have finished this proof.

02:33 Now for the part b, what we need to do is what happened, well, basically the statement say that if f and g are onto, then the composition is also onto.

02:52 What the meaning of onto is that, in this case, g -o -f is a map from a to c.

03:02 We need to prove that for all the elements on the set c, there exist an element a on a, such that g -o -f of a is equals to c.

03:20 Okay, so we need to prove this.

03:24 How to start.

03:25 I'm going to use some diagrams to help with the intuition of this proof.

03:32 We got this.

03:39 You got here all the elements of c.

03:42 Let's suppose that you take any, without any restriction.

03:47 And we know that g is a map from b to c.

03:53 Moreover, we know that g is onto.

03:57 Because g is onto, that means that for all the elements on c, there exists an element on b, okay? there exists an element on b such that g of b is equal to c...

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