# Tag Info

## Hot answers tagged homotopy-theory

42

But after all, $\pi_n(X,*)$ is "merely" the set of morphisms in the homotopy category of pointed spaces from the $n$-sphere to $X$: $\pi_n(X,*) = \hom_{\mathsf{hTop}_*}((S^n,*), (X, *))$. This has always seemed somewhat "biased" to me; spheres are given a preponderant role in the definition. Spheres emerge naturally out of the relationship between homotopy ...

40

The following argument is essentially an application of the path lifting property for covering spaces. Let's think about $\mathbb{R}P^2$ as being the quotient space you get by identifying antipodal points on the sphere $S^2$. That is, let $x\sim -x$, let $\mathbb{R}P^2=S^2/\sim$ and let $p\colon S^2\rightarrow\mathbb{R}P^2$ be the quotient map. Let $z$ be ...

21

I believe there is a nontrivial map $\mathbb{C}P^k\rightarrow \mathbb{C}P^{k-1}$ when $k > 2$ is odd. I don't know what happens when $k$ is even. Consider the following composition $$S^{2k+1}\rightarrow \mathbb{C}P^k\rightarrow S^{2k}\rightarrow S^{2k-1}\rightarrow \mathbb{C}P^{k-1}$$ with $k$ odd, where I'll now describe all the maps. The maps from ...

20

The identity is not homotopic to a constant map; otherwise, $S^1$ would be contractible, which would imply $\pi_1(S^1)=0$.

18

No. In fact, the homotopy groups of the wedge sum are often vastly more complicated. For example, consider $S^1\vee S^2$. Its universal cover looks like a line with an $S^2$ wedged at every integer point. In particular, $\pi_2$ of the universal cover is infinitely generated (one map for each of the countably many spheres), despite the fact that $\pi_2(S^... 18 I can prove that there at most$2$homotopy classes of maps from$\mathbb{C}P^2$to$\mathbb{C}P^1$. However, I can't prove or disprove that there are exactly$2$. Perhaps someone can answer the final question and finish this off? (Or provide a completely different proof!) I'll use the notation$[X,Y]$to refer to homotopy classes of maps from$X$to$Y$.... 17$f : S^1 \to S^1$be a map with no fixed points. Consider the projection of the straightline homotopy $$H(s, t) = \frac{(1-t)f(s) - ts}{\left \lVert(1-t)f(s)-ts\right \rVert}$$ between$f$and the antipodal map$-\text{id}$, which is well-defined since$f(x) \neq x$for all$x$. Compose this with the homotopy $$H(s, t) = e^{i\pi (1-t)} s$$ which rotates$-\...

17

Perhaps the equivalence relation you want is "isotopic", which briefly means "homotopic through homeomorphisms". More precisely, you want to know whether there is an isotopy from $C_0$ to $C_2$, which by definition is a continuous function $H : C_0 \times [0,1] \to \mathbb{R}^3$ such that $H(x,0)=x$, $H | C_0 \times \{t\}$ is a homeomorphism onto its image ...

17

You are correct. This follows from the fact that $U(N)$ is diffeomorphic to $S^1\times SU(N)$. To see this, consider the map $f : U(N) \to S^1\times SU(N)$ given by $A \mapsto (\det A, \operatorname{diag}((\det A)^{-1}, 1, \dots, 1)A)$. This is a smooth map with smooth inverse given by $(z, B) \mapsto \operatorname{diag}(z, 1, \dots, 1)B$. Note however that ...

16

In general, finding the isomorphism class of higher homotopy groups of the spheres is an extremely hard problem. Luckily however, there seem to be some techniques available to the low dimensional spheres which aren't available to the higher dimensional spheres. In particular, there appears to be an isomorphism between $\pi_n(S^2)$ and the co-kernel of a ...

16

One person offering three answers to one post seems ludicrous (and this is a first for me), but I've really enjoyed this problem. Theorem: There is a homotopically nontrivial map $\mathbb{C}P^2\rightarrow S^2$. (By my second post, this map must be unique up to homotopy.) We will prove this by answering the question in post 2. Namely, we show that ...

16

That statement is true. First, every free group $F$ is a the fundamental group of a bouquet of circles (having a common base point); each circle represents a generator. Now every group $G$ is a quotient of a free group $F$ by a subgroup (of $F$) generated by the "relations" $f_i\in F$. Note that each $f_i$ is a word in $F$, that is, a product of the ...

15

A topological space is a pair of a set and a family of subsets (closed under certain operations that give to such family the structure of a topology). Homemorphisms are exactly the more natural maps between such spaces: given two topological spaces $\langle X, \tau_X \rangle$ and $\langle Y,\tau_Y \rangle$ an homemorphisms $f \colon \langle X,\tau_X \rangle ... 15 The focus on weak equivalences instead of homotopies is largely a consequence of Grothendieck's slogan to work in a nice category with bad (overly general) objects, rather than working in a bad category that has only the good objects. Typically, there is a good notion of homotopies between maps that is well-behaved, but only on the "good objects". If we ... 14 Since you are now asking for a connected example, here is a justification for Harry's example: Lemma. The Hawaiian earring$H$is not homotopy-equivalent to any CW-complex$X$. Proof. Suppose to the contrary,$f: H\to X$is a homotopy-equivalence. Since$H$is compact, so is$f(H)$. Hence,$f(H)$is contained in a finite subcomplex$Y$of$X$. In ... 14 You can take a look at Robert Harper's lectures: http://www.cs.cmu.edu/~rwh/courses/hott/ (There are lecture notes and video recordings) They require much less than you described to understand and covert HoTT. 14 In case we do wish to fix a basepoint on$\Sigma$: Consider a torus$\Sigma=T^2$in$E^3$, and let$U$be the closure of the bounded region of$E^3-\Sigma$. One option is that$U$is a solid torus, in which case the induced map$\pi_1(\Sigma)\to \pi_1(U)$is surjective. The other option is that$U$is a nontrivial knot complement. For example, the ... 13 You cannot say that$X$and$Y$are homotopy equivalent if$\pi_i(X) = \pi_i(Y)$for all$i$. This is because the definition of homotopy equivalence requires that there is some continuous map$f: X \longrightarrow Y$that is invertible up to homotopy. Hence the isomorphisms$\pi_i(X) \cong \pi_i(Y)$need to be induced by such a homotopy equivalence$f$. The ... 13 This result is proved in a second paper by Arne StrÃ¸m, "Note on Cofibrations II" in Math. Scand. 22 (1968), 130-142. As far as I am aware this is the original source, and in any case it was the source for the proof given in the appendix of my book (added in 2009 to the online version of the book). I will add this paper to the list of references in the book -... 13 I will add my answer here to give the readers of this question a bit more explanation as to how one could prove this. Proof: Suppose that$X$is contractible. Choose any two points$a, b \in X$. We will construct a path between$a$and$b$. Since$X$is contractible, the identity map$\operatorname{id}_{X} : X \to X$defined by$\operatorname{id}_X(x) = x$... 13 The usual condition says that a map$\Lambda^k[n+1] \to X$can be extended to$\Delta[n+1] \to X$. Your other condition says that a map$\partial\Delta[n+1] \to X$can be extended to$\Delta[n+1] \to X$. The latter condition implies the first condition, but even more so, it implies that$X$is contractible. One can see this by considering geometric ... 13 Yes. Actually there is a more general result. If$X=\bigcup_{i=1}^n A_i$where each subspace$A_i$is contractible, then the product of any$n$elements in$H^*(X)$vanishes. Since$\Sigma X$is a union of two cones, each of which is contractible, the result follows. To establish the general result, simply note that$H^*(X)\cong H^*(X, A_i)$as$A_i$is ... 13 As soon as you have a homotopy invariant functor, it factors through the homotopy category, by the universal property of localizations. To give a homotopy invariant functor$\mathsf{C} \to \text{Whatever}$is exactly the same thing as giving a functor$\operatorname{Ho}(\mathsf{C}) \to \text{Whatever}$. So if we can understand the homotopy category well ... 12 Another way to look at those notions of groupoids and fibrations is that type theory tells you what they are. It does so by telling you what you can do with them. For example with fibrations, you can transport points in the fibers along paths -- this is exactly the intention of the classical notion of fibrations. It is even more true for weak$\infty$-... 12 Yes, there is a spectral sequence for which the groups on the$E^1$-page are the singular homology of$D(i)$for varying$i \in I$, which converges to the singular homology of$X$. This spectral sequence arises from the skeletal filtration$X_n$of the homotopy colimit$X$: essentially,$X_n$is the part of the hocolim that comes from$k$-tuples of arrows in ... 12 No. In general, homotopy groups behave nicely under homotopy pull-backs (e.g., fibrations and products), but not homotopy push-outs (e.g., cofibrations and wedges). Homology is the opposite. For a specific example, consider the case of the fundamental group. The Seifert-Van Kampen theorem implies that$\pi_1(A\vee B)$is isomorphic to the free product$\...

12

Observe that any map $\mathbb{CP}^{n} \rightarrow \mathbb{CP}^{\infty}$ can be pushed down to the $2n$-skeleton of $\mathbb{CP}^{\infty}$, due to cellular approximation. This establishes a bijection between homotopy classes of maps $[\mathbb{CP}^{n}, \mathbb{CP}^{\infty}] \simeq [\mathbb{CP}^{n}, \mathbb{CP}^{m}]$ for all $m \geq n$. The left hand side ...

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