Amitesh Datta
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 Jun28 comment How to study math to really understand it and have a healthy lifestyle with free time? Hi @JPG, thanks for your comment! It's hard for me to recommend anything without knowing your specific situation but I think the main thing is to think about mathematics in your own way, in addition to what you are learning. Of course, there are many ways of doing this and what I suggested is just one way. It's worth trying various approaches and then deciding which ones work for you. Jun27 comment How to study math to really understand it and have a healthy lifestyle with free time? Hi @Pacerier, I wrote them in several notebooks by hand, and they are probably somewhere in my house. Perhaps I will try to find them now that you mention it! Jun22 comment How to study math to really understand it and have a healthy lifestyle with free time? Hi @Pacerier, thanks for your comment. I've never actually gone back and re-read my notes (though I would be curious to do so at some point, even if only to see what I was thinking the first time I was learning the subject). If I do review a subject, then for the reason you mention, I would go back and re-read the relevant parts of the book (which would probably be better exposition then my rough notes, anyway). Nowadays, I don't adopt this learning style; it takes too much time and doesn't necessarily help me to remember/understand the subject better! Jun18 comment Show that there is no analytic bijection from the unit disc to $\mathbb{C}$ Hi @Aloysius, if I understand your question correctly: "set-theoretically", the map admits an inverse because it is bijective. Now, we know that locally this inverse map is holomorphic, so it must therefore be (globally) holomorphic. Jun18 comment Show that there is no analytic bijection from the unit disc to $\mathbb{C}$ I guess most complex analysis textbooks should contain this fact. However, here is a proof sketch ($\phi$ is holomorphic near $0$ and WLOG $\phi(0)=0$): if $\phi'(0)\neq 0$, then $\phi$ admits an analytic inverse near $0$; otherwise, $\phi$ has a zero of order $n>1$ at $0$ and $\phi(z)=z^n\psi(z)$ for some $\psi$ which does not vanish near $0$. Locally, $\psi$ thus has an $n$th root (say, $\eta$) so we can write $\phi(z)=(z\eta(z))^n$ which is clearly not injective (it is $n$-to-$1$). I've possibly used something non-trivial in this proof, so let me know if I should elaborate. Jun18 comment Show that there is no analytic bijection from the unit disc to $\mathbb{C}$ Hi @Aloysius, it's a special property of holomorphic functions that a bijective holomorphic function admits a holomorphic inverse (an analogous statement is not true for real analytic functions as $f(x)=x^3$ shows). The key fact needed is that an injective holomorphic function must have non-vanishing derivative. Are you aware of this? Jun13 comment About the topology textbook. I second the suggestion of @alkabary of using Munkres (this was the book that furnished my first introduction to topology). Jun3 answered The galois group of a polynomial of degree 3 is either $A_3$ or $S_3$ Jun2 comment section of a fiber bundle In fact, all of the bundles mentioned in the answers (so far) are non-trivial principal bundles. :) Jun2 comment section of a fiber bundle +1 :) Just a note for people who read both answers, Slade's answer is a great explanation of the fact that the double cover $S^1\to \mathbb{RP}^1$ (the case $n=1$ of $S^n\to\mathbb{RP}^n$ mentioned in my answer) does not admit a section. (I see that Stefan has mentioned this particular example as a comment to the question and Lee has also mentioned this example as a comment to my answer.) Jun2 answered section of a fiber bundle May31 comment Proving that evry polynomial of odd degree has at least one root on R Let $p$ be a polynomial of odd degree with real coefficients. Evaluate $\lim_{x\to\infty} p(x)$ and $\lim_{x\to -\infty} p(x)$. Then, apply the intermediate value theorem. The theorem will not (in some sense) admit a purely algebraic proof because it is not true for polynomials with rational coefficients (restricted to the rational numbers); we need to use the (essentially, analytic) construction of $\mathbb{R}$ from $\mathbb{Q}$. May30 comment Next book in learning Differential Geometry Hi @AlphaE, I read Boothby's book (that's where I first learnt about differentiable manifolds); I thought it was quite a well-written book. (I'd be curious to know why you think otherwise.) I think do Carmo summarizes a lot of the elementary material that he needs (much of which would be covered in more detail in Boothby's book, for example) in Chapter 0. I don't know how practical it would be to learn this material directly from Chapter 0 of do Carmo's book, though; it depends on your mathematical maturity. May30 comment Next book in learning Differential Geometry Hi AlphaE, yes that's it (I wrote the full title in my previous comment). By the way, as @littleO sort of suggested, there are a number of directions other than differential geometry which you could take. Lee's book is (if I remember correctly) on the general theory of topological manifolds and probably covers some algebraic topology. So, if you are interested in algebraic topology, you could read that, or a number of other references too. (Actually, I think that knowledge of the fundamental group and differential forms/de Rham cohomology is probably necessary for "Riemannian Geometry" too.) May30 comment Next book in learning Differential Geometry What about do Carmo's "Riemannian Geometry" (which is, in some sense, a sequel)? The book covers some of the foundational material in Riemannian geometry that you would need to study modern Riemannian geometry and research papers in the field. After, that there are a number of possible directions you could take, which I would be happy to note if you are interested. May29 comment Characterization of 1-dimensional manifolds. Yes, it's true (assuming second countable and Hausdorff too, to exclude the long line and the line with a double origin, respectively). I think this has been asked before: math.stackexchange.com/questions/113705/… May28 comment Show that $\mathbb{R}^m$ is not homeomorphic to $\mathbb{R}^n$ Hi @Ant, no, for example, $\mathbb{R}^m$ is noncompact, yet $\mathbb{S}^m$ is compact. (Of course, there are other proofs too.) May27 comment Is it useful to learn math by proving a formula/theorem? Hi @St.ClairBij, regarding self-study of mathematics, I think there are numerous questions and answers already on this website on the topic. You might be interested in click me for example. A complete list of questions with the tag self-learning is available at click me. Finally, there are numerous questions on this topic listed under "Related" on the right-hand sidebar. May27 comment Is it useful to learn math by proving a formula/theorem? I voted to close this question because I think it's too broad and primarily opinion based. Do you have a particular branch of mathematics in mind? I don't think most branches of mathematics admit a list of theorems which are universally considered to be fundamental, especially more advanced or recent branches of mathematics. I also don't know if it is meaningful to break up mathematics into isolated formulas/theorems; mathematics is a story and how you tell that story matters. May27 comment real valued functions with composition Hi Joel, I deleted my first comment because I saw you were editing, so I thought it might not be relevant anymore. Anyway, +1 from me. :) @MalJA the key point is that $G$ does not have any identity, as the only possible identity would have to be the identity function, which does not always assume nonnegative values.