Let $A = C^\infty(S^1)$ be the ring of smooth functions on the circle (if you prefer, you can see it as the ring of smooth $2\pi$-periodic functions $\mathbb R \to \mathbb R$). First, $A$ isn't ...

I'll just add a remark to Georges's (excellent) answer. The real plane $\mathbb R^2$ has two freely substitutable orientations but the complex plane $\mathbb C$ has a canonical one (as a real vector ...

I'm quite late on this one, but I think the result is nice enough to be included here. Definition A function $f : \mathbb R \to \mathbb R$ is said to be convergence-preserving (hereafter CP) if $\sum ... View answer Accepted answer 30 votes This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof. The proof I'm ... View answer 19 votes Here's an article (in French) that aims to explain the mathematics behind the game to a wide audience. In the interest of link rot prevention, here are two diagrams from the article that may be of ... View answer Accepted answer 12 votes There is a map$\epsilon : C_0(X) \to \mathbb Z$. It is indeed a part of the complex for reduced homology, but it exists independently. Because$\epsilon \partial_1$is zero, the value of$\epsilon(z)$... View answer 11 votes OK, my bad, Fulton's Algebraic topology: A First Course only deals with the closed case. I'll suppose that you know this case quite well. Let's do the bounded case by hand. First case: one boundary ... View answer Accepted answer 11 votes I will denote by$LH^*$the free part of the cohomology ring. (That means that$LH^r(M)$is$H^r(M)/\mathrm{torsion part}$.) For a connected 4-manifold, the symmetric form$I : LH^2(M) \times LH^2(M) \...

Yes. In general, $N \subset M$ is a submanifold if you can find for every $x \in N$ an open neighbourhood $U \subset M$ of $x$, an open neighbourhood $V \subset \mathbb R^{\dim M}$ of $0$ and a ...

Just a “visual” construction of the isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$... It's quite well known that $4 = 2 +2$. Concretely, it means that if you have four things, you can ...

Yes. This procedure is called clutching (and the resulting spheres are clutched spheres or twisted spheres In this procedure $g$ is a diffeomorphism. If $g$ extends to a diffeomorphism of the whole ...

Look at the classification of covering spaces (e.g. Hatcher's Theorem 1.38 and Proposition 1.39): A two-sheeted cover $p : (\tilde X, \tilde x_0) \to(X, x_0)$ is determined by the subgroup $\Gamma = ... View answer Accepted answer 9 votes What seems to be clear is that the origin of the Grothendieck group is Grothendieck's work on Riemann-Roch's theorem around 1956. According to Weibel's The Development of Algebraic K-theory before ... View answer 9 votes You have to understand that the notion of function as it is used nowadays is quite recent. During a long time, analysts were perfectly happy to work with so-called multiform functions. For example,$\...

If $A$ is a commutative ring, a classical result states that the polynomial ring $A[x]$ is a PID if and only if $A$ is a field. It is a good exercise. In your case, as $F[x]$ isn't a field, $F[x,y] \... View answer Accepted answer 8 votes Here's the example I give to my students: there's a rule that says that "If a traffic light is red, then you must stop". Recall that this is what implication means:$P \implies Q$means "if$P$, then$...

In On Numbers and Games, Conway defined a field structure on the set of all ordinals, and he calls the result $\mathbf{On}_2$. It is an algebraically closed field of characteristic two, if you are ...

You're mostly right. The relation between complex structures and metrics comes from their common passion about angles. Basically, a complex structure on a Riemann surface is just a procedure for ...

Your question can be subject to interpretation, but here's a proof that you cannot hope for too strong a result. Take the simplest example you can think of: $M'$ is the $n$-ball, and you'll add a ...

It's a famous open problem : the perfect cuboid problem.

Hansjörg Geiges's Introduction to Contact Topology seems to be the only textbook-style reference on Contact Geometry. (At least it was three years ago, but I'm unaware of a more recent book with this ...

It is not true that if $A$ is not finitely generated, you can find $\mathbb Z$-independent elements. For example, $\mathbb Q$ is a non finitely generated $\mathbb Z$-module, but, clearly, you cannot ...

Local coefficient homology is a particular case of sheaf cohomology (cohomology of a locally constant sheaf). So, even if I'm not sure I understood precisely your wishes, I think it is possible that ...

Let $$g : \begin{array}{ccc} \mathbb R & \to& \mathbb R\\ c &\mapsto &f(c+\pi) - f(c).\end{array}$$ That's a continuous function. Now, let $c_0$ be a point where $f_{|[0,1]}$ has a ...

It is not true that Aut(X) acts transitively. It's for example false for all Riemann surfaces of genus > 1, as the automorphism group is then finite.

Let's listen to the master: Mais en 1945, Jacobson observe [172 c] que le procédé de définition d'une topologie, imaginé par Stone, peut en fait s'appliquer à tout anneau A (commutatif ou non) pourvu ...

Proposition 3 is false, or at least phrased in a misleading way: the complement of a knotted solid torus is S³ is certainly not a solid torus: this happens if and only if the solid torus is unknotted, ...

No, because of the Poincaré-Volterra theorem. Here's the statement you can find in Bourbaki's General Topology (I.11.7, corollary 3). Theorem. Let $Y$ be a locally compact, locally connected space ...

The category of affine schemes is nothing but the opposite category of commutative rings. So $\mathrm{Spec}\, \mathbb Z$ is a terminal object in the category of affine schemes because $\mathbb Z$ is ...