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Jan
31
comment Defining natural transformations based on generalized elements?
As an additional remark, yoneda also implies that the natural isomorphism $\langle -,-\rangle$ determines completely the structure of the product on $A \times B$.
Jan
31
answered Defining natural transformations based on generalized elements?
Jan
28
awarded  Nice Answer
Jan
28
comment If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
Finally if you look carefully to the counterexample above that exactly what it does: it takes two unrelated natural transformation between the identity functors, i.e. two elements of the set $\mathbb C^{\mathbb C}[1_{\mathbb C},1_{\mathbb C}]$. To prove that the functor $1_{\mathbb C}$ is self-equivalent we just need to provide an element of that set, this could be the identity but it doesn't have to be it.
Jan
28
comment If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
I think that the point is that while we can choose many different, unrelated, proofs of the fact that $F$ and $G$ are equivalences, if we want that our proofs also .... prove that the functors are adjoint we need to choose them more carefully.
Jan
28
comment If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
@StudentType think of the natural isos as proofs of the equivalences between the functors. An equivalence of categories provides two functors between the categories with two proofs that these functor are equivalences of categories. In an adjoint equivalence we need to require something more, for instance we want that the two proofs satisfy the triangle identities.
Jan
27
revised Do mathematical objects have underlying types?
Left unrelated tags
Jan
27
answered Quantifier notation: $\forall n \implies \cdot$ versus $\forall n, \cdot$
Jan
27
revised If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
added 6 characters in body
Jan
27
comment If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
By the way, it is true that if $\langle F,G,\alpha,\beta \rangle$ is an adjoint equivalence then $\alpha_{G(D)}=G(\beta_D)$ for every $D \in \mathbf D$, that's a consequence of the triangle identities.
Jan
27
revised If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
Make a correction
Jan
27
answered If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
Jan
26
answered How do you prove that $Y^*(1_B) = 1_{Y^*B}$ given $Y^*(f) = \mathcal{A}(f, -)$?
Jan
24
answered Homotopy equivalent but not deformation retraction
Jan
24
revised Homotopy equivalent but not deformation retraction
corrected a typo
Jan
24
comment Is the axiom $g1 = g$ essential for a group action
Just a remark: how would you conclude that $(\omega\cdot g)\cdot g^{-1}=\omega$? Usually to prove that one uses the axiom $\omega \cdot 1=\omega$.....
Jan
24
answered Categorical introduction to Algebra and Topology
Jan
23
answered Homotopy colimit,weighted colimit, homotopy theory
Jan
23
comment Isomorphism between categories and how to prove non-isomorphism
I see so basically is the generalization of the composition of two anti-monotone maps.... to be fair I was expecting that but I wanted to be sure, thanks. :)
Jan
23
comment Isomorphism between categories and how to prove non-isomorphism
Just a question: if there is a category whose morphisms include contravariant functors how do you compose two contravariant functors between them? More in details: if you have $F \colon C \to D$ and $G \colon D \to E$ are two contravariant functors (so $F: \colon C^\text{op} \to D$ and $G \colon D^\text{op} \to E$) what is $G \circ F$?