# Omar Antolín-Camarena

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bio website math.harvard.edu/~oantolin location Cambridge, Massachusetts, USA age 33 member for 3 years, 7 months seen 2 days ago profile views 272

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 Dec10 comment Values of a, b, and c that the curve $y = ax^3 + 3x^2 + bx + cx + e^x$ has one point of inflection? You don't need to solve $6ax+6+e^x=0$, @DinoMint, only figure out how many solutions it has. Think of it as intersecting the graph of $y=e^x$ with the line $y=-6ax-6$. If the slope $-6a$ is negative, there is exactly one intersection point, if it is positive you can find the tangent line to $y=e^x$ with the same slope ($-6a$) and see if this line, $y=-6ax-6$ is above or below the tangent. Dec6 comment Is an equivalence an adjunction? @MartinBrandenburg: it can be characterized as either the unique book with title "Homotopy Type Theory" or as the unique book on the subject of Homotopy Type Theory. Nov18 comment Direct limit and products Usually, "... doesn't hold in any ..." means "never holds", @MartinBrandenburg, you probably meant to say "doesn't hold in every category", since it does hold in some, such as the category of sets which is the one Sean said he was interested in. Nov13 comment Direct limit and products The StackExchange software noticed my answer was trivial and converted it to a comment. Cool. I'm surprised it hasn't noticed how trivial some of my other answers are. Nov13 comment Direct limit and products Steve Lack wrote a proof in this MathOverflow question of his. Nov8 comment How to compute the fundamental group of a necklace of $\mathbb{S}^1$' s? Hatcher's Algebraic Topology, section 1.A, page 83. Nov8 answered How to compute the fundamental group of a necklace of $\mathbb{S}^1$' s? Nov7 comment Associativity of a composition $x ∘ y = xy+\sqrt{(x^2-1)(y^2-1)}$ Doesn't using $\cos$ instead of $\cosh$ show the second operation is associative in exactly the same way? (At least on $[-1,1]$.) Nov1 revised How to prove exactness implies complex? deleted 110 characters in body Nov1 answered Category equivalence of sets and vector spaces Nov1 answered How to prove exactness implies complex? Nov1 comment Compute homotopy classes of maps $[T^{2},T^{2}]$ This answer shows that $[T^2,T^2] \cong \mathbb{Z}^4$, and it is not hard to show that a representative of the class $(a,b,c,d) \in \mathbb{Z}^4$ is given by the function $T^2 \to T^2$, $(w,z) \to (w^az^b, w^cz^d)$ (where I think of $T^2$ as $S^1 \times S^1$, and of $S^1$ as the unit norm complex numbers). That map has degree $ab+cd$, by counting pre-images. Therefore the map $deg : \mathbb{Z}^4 \cong [T^2, T^2] \to \mathbb{Z}$ is given by $(a,b,c,d) \mapsto ab+cd$, @R.Bradley. Oct29 comment Working with $\bigotimes_{\mathbb{Z}_X}$? What kind of objects are the entries of $B$? What do you mean by "$A$ is a homology"? Oct28 comment Do adjoint functors really define monads? I'd just add that the category of $\mathfrak{g}$-modules is a category of left modules for an associative algebra, but that associative algebra is the universal enveloping algebra $U(\mathfrak{g})$ (not $\mathfrak{g}$, which, as you point out, is not associative). Oct28 comment Do adjoint functors really define monads? @GiorgioMossa: He was wrong, the left adjoint to the forgetful functor is $V \mapsto U(\mathfrak{g}) \otimes V$, so it cannot also be $V \mapsto \mathfrak{g} \otimes V$ (a left adjoint is uniquely determined up to natural isomorphism by the right adjoint). Oct27 comment Do adjoint functors really define monads? Oh, sorry, I only posted the first comment because I believed you were right in saying that the identity was false! But now I'll ask something else: are you sure the free $\mathfrak{g}$-module on $V$ is $\mathfrak{g} \otimes V$? Because the definition of a $\mathfrak{g}$-module is rigged to match that of a module over the universal enveloping algebra $U(\mathfrak{g})$, I would expect the free module to be $U(\mathfrak{g}) \otimes V$. Oct27 comment Do adjoint functors really define monads? @AlešBizjak: you're right of course, I don't know what I was thinking. Oct27 comment Do adjoint functors really define monads? Why do you think $\epsilon\circ (LR\epsilon)=\epsilon\circ(\epsilon LR)$ should hold? If your are explicit about this people can probably pinpoint your confusion. In the monad from an adjunction the multiplication is $R \epsilon L$, so the monad laws you need to check will involve that and not naked $\epsilon$'s or the combination $LR\epsilon$. Oct15 revised A confusion about the fact that contractible spaces are simply connected added 793 characters in body Oct15 answered A confusion about the fact that contractible spaces are simply connected