Giorgio Mossa
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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 $G \circ F$?