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Let $G$ be a topological group with classifying space $BG$. Suppose that $BG$ supports a group structure. Then there are homotopy equivalences $G\simeq \Omega BG\simeq\Omega^2B^2G$, so $G$ is a double loop space, and in particular supports a homotopy abelian H-space multiplication. However: Theorem: Let $X$ be a non contractible, connected, finite complex ...


4

If $f$ is surjective then $f(S^1)=S^1$ and thus $\pi_1(f(S^1))=\pi_1(S^1)\simeq\mathbb{Z}$. On the other hand if $f$ is not surjective, then the image $f(S^1)$ is a subset of $S^1\backslash\{p\}$ for some point $p\in S^1$. And the latter space is homeomorphic to $\mathbb{R}$. And thus $\pi_1(f(S^1))$ has to be trivial, since path connected subsets of $\...


4

I mean... Off the top of your head do you see what the cohomology of $\mathbb{Z}$ should be? Now ask yourself if you can see immediately what the cohomology of $S^1$ is. For me, at least, one of these questions is much easier. More broadly, what's the cohomology of a free group $F_k$? What about a wedge of $k$ circles? These examples are slightly silly ...


4

Since you mention the Atiyah-Hirzebruch spectral sequence, I want to point out how you might construct a (minimal) example if you somehow didn't know any examples of topological spaces (but do know quite a bit of algebraic topology). As we're working with the $K$-theory of spaces, we would like to understand the structure of the connective complex $K$-theory ...


4

Your ideas are correct, but you have to make explicit how you regard an element of $\mathbb Z^{\mathbb N}$ as an element of $\pi_1(H)$ and that the resulting function $\phi : \mathbb Z^{\mathbb N} \to \pi_1(H)$ is injective. Let us elaborate 1. Let us write $l_n^m : [0,1] \to K_n$ for the loop based at $0$ that travels counter-clockwise $m$ times around $...


4

Here is one reason it turns out to be useful. I won't attempt to be exhaustive, and hopefully other people will also contribute some more ideas. By construction homotopy is a much coarser invariant than homeomorphism: two spaces which are homeomorphic are trivially homotopy equivalent. On the other hand homotopy does not even preserve basic topological ...


3

The short answer to your main question is "no". If $A_*$, $B_*$, and $C_*$ are chain complexes, having $C_q = A_q \oplus B_q$ for each $q$ in some sense doesn't guarantee that $C_* = A_* \oplus B_*$ as chain complexes. The relevant maps between $C_q$ and $A_q \oplus B_q$ need to also respect the boundary maps in the chain complexes in order for ...


3

Sometimes one can. For instance, if $H$ is an abelian subgroup: If the subgroup is cyclic, you can, if the subgroup is $Z^2$ there is a precise condition when you can (it is called the "torus theorem"). One can sometimes find reasonable conditions for this when $H$ is the fundamental group of higher genus surface. This works better if the manifold $...


2

Tyrone: Plenty is known about the case you suggest. The space of long knots just removes some ambiguity. Remember that there is a fiber sequence $\text{Emb}_*(S^1, S^3) \to \text{Emb}(S^1, S^3) \to (TS^3 \setminus 0)$, the last map sending $\gamma$ to $\gamma'(0)$ and the first is the space of embeddings with prescribed values for $\gamma(0)$ and the ...


2

For complex line bundles $L$ the first Chern class $c_1(L)$ takes values in $H^2(-, \mathbb{Z})$ and this is quite specific to working over $\mathbb{C}$. One way of thinking about this class is as coming from the exponential sequence $$1 \to \mathbb{Z} \xrightarrow{2\pi i} \mathbb{C} \xrightarrow{\exp} \mathbb{C}^{\times} \to 1$$ which induces a long exact ...


2

This is too long to post as a comment, so I'm leaving it here instead. I'm not sure how rigorous an argument you're looking for. Truly rigorous arguments in Algebraic Topology are really hard to write, and are also quite hard to find in the literature. (I have some opinions on this, which I'm putting at the end). I'll say this, though: If I were grading an ...


2

Let call $S$ to your set. CLAIM: Any point in $S$ can be arc-connected with $(2,0,0)$. PROOOF: Let $u\in S$, then the segment from $u$ to $\frac{2u}{|u|}$ always lies in $S$. Now take any continuous path form $\frac{2u}{|u|}$ to $(2,0,0)$ lying in the sphere centered at $(0,0,0)$ with radius $2$ (this path also lies in $S$). Join the two paths and you're ...


2

The examples that you have worked out exhibit a pattern, of which you have stated some special cases. Here's the general pattern: Proposition: If $X$ has an abelian fundamental group $\pi_1(X)$, then $X$ is homeomorphic to the quotient of a circle $C$ and a finite collection of trees $T_1,...,T_k$, by choosing one vertex $x_i \in T_i$ on each tree and a ...


2

A more common notation for a solid space is "absolute extensor for normal spaces". Your construction of $f'$ shows that $(Y,y_0)$ is pointed contractible for each $y_0 \in Y$. This immediately implies that For each open neigborhood $U$ of $y_0$ in $Y$ there exists an open neighborhood $V$ of $y_0$ in $Y$ contained in $U% $ such that the inclusion $...


2

Hint: if a group has a proper subgroup of finite index then it has a proper normal subgroup of finite index. To see this, let $H$ be a proper subgroup of $G$ with $n = (G:H)$ finite. The action of $G$ on the left cosets of $H$ induces a homorphism of $G$ into a non-trivial subgroup of $S_n$. The kernel of this homomorphism is a proper normal subgroup of ...


2

Here is a simple example which exhibits that the answers to your questions in 1 are "no". Let $Y = I$, and let $U_\alpha$, for each point $\alpha = (y,t) \in Y \times I$, be the box neighborhood $$U_\alpha = (I \times I) \cap \bigl((y-.001,y+.001) \times (t-.001,t+.001)\bigr) $$ You can see, for example, that its going to take a few hundred ...


2

Let $Q$ be the quadratic form on $\mathbb{R}^{n+1}$ defined by $$ Q(x_0,\dots,x_n)=\sum_{i=1}^{n+1}\alpha_ix_i^2,$$ where $0<\alpha_1<\dots<\alpha_{n+1}$ is a sequence of $n$ positive real numbers. Now take the standard embedding $S^n\subset\mathbb{R}^{n+1}$ of the $n$-sphere and let $\widetilde f=Q|_{S^n}:S^n\rightarrow\mathbb{R}$ the restriction ...


2

Good example is $X=\mathbb RP^n$ (see e.g. $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$).


1

Actually, the statement is true if and only if $Y$ is path connected. For the second direction, suppose any two continuous maps $f,g: X\to Y$ are homotopic. Take any two points $y_0, y_1\in Y$, and we will show there is a path in $Y$ which connects them. By assumption, the constant maps $f,g:X\to Y$ defined by $f(x)=y_0, g(x)=y_1$ are homotopic. Thus, there ...


1

First of all, you are trying to prove a false result. Try to draw $S$: it is the exterior of the unit ball and it is clearly "visually-connected". As connectedness has been invented to give a rigorous definition of "visually-connected", this would be a strange fact if $S$ was not connected. To prove $S$ is connected, you can show: every ...


1

You're not missing anything that comes to my mind, the only thing you should be aware of is that these classes come with a natural structure of group, given by the "composition" of loops. This group structure however is generally not commutative: in your case, $\mathbb R^2$ with two holes makes space for all possible composition of loops around ...


1

The concern you mentioned at the end of your post is pretty accurate. It may be helpful to recall how you arrived at the fundamental group of your covering space, this should give you an answer to your question. The computation can be done via the Seifert-Van-Kampten theorem, that inductively gives you the fundamental group your wedge sum. But to make life ...


1

Let $p : \tilde X \to X$ be any covering map and $f : Y \to X$ be any map. It is a well-known theorem in the theory of covering maps that for any two lifts $\tilde {f_1}, \tilde {f_2}: Y \to \tilde X$ of $f$ the set of points of $Y$ where $\tilde {f_1}$ and $\tilde {f_2}$ agree is open and closed in $Y$. See for example Hatcher's book on Algebraic Topology. ...


1

Yes, this is true in general. Suppose that $X$ is a wedge of $k$ circles so that its fundamental group $F_k$ is free of rank $k$. Orient each loop. These loops, $x_1, \dots, x_k$, define a basis. Suppose we are given a group $G$ (such as $A_4$) and a surjective homomorphism $f:F_k \to G$. This defines a generating set for $G$, namely $f(x_1), \dots, f(...


1

Let's say that we're given a map $g: S^1 \rightarrow X$ and we want to know when it extends to a map $D^2 \rightarrow X$. This happens precisely when $g$ is nullhomotopic. If $g$ is nullhomotopic then there is a homotopy $H:S^1 \times I \rightarrow X$ so that $H(-,0) = g$ and $H(-,1)$ is a constant map. This means that $H$ descends to a map $H':(S^1 \times I)...


1

As a classifying space, the Grassmannian $\mathrm{Gr}_m(\mathbb{C}^\infty)$ can be modelled as the quotient of a contractible space with free $U(m)$-action by the group $U(m)$. Therefore, in order to show that $$\frac{U(\mathcal{H})}{U(V) \times U(V^\perp)}$$ is a Grassmannian, it suffices to show that the Stiefel manifold $\mathrm{St}_m(\mathcal{H}) := U(\...


1

First, I'll note that homotopy groups do not even solely take values in groups since $\pi_0$ is a set. Even if we ignore $\pi_0$ they do not take values in abelian groups since $\pi_1$ may be noncommutative. Let's suppose we ignore both of those issues. If the homotopy groups were a generalized homology theory, then they would have a suspension isomorphism. ...


1

Look at chapter 9 (especially Theorem 9.12) of Switzer, Robert M. Algebraic Topology - Homotopy and Homology. Springer, 2017 or at chapter 7 (especially Theorem 7.7.14) of Spanier, Edwin H. Algebraic topology. Springer Science & Business Media, 1989.


1

Alright, with the help of Eric Wofsey's hints, I think I have an answer to my question. First we have that $H^k(T,S^1) \cong \tilde{H}^k (T/S^1)$ for all $k$, where the quotient map is induced from the embedding of the circle. (I cannot find a reference for this fact.) Note that whenever a (nice enough?) space is path connected, the $0$th reduced homology is ...


1

Compactness of a set is a topological property: it is independent of the space in which that set is embedded. $A$ is a compact subset of $\Bbb R^2\setminus\{\bf0\}$, so it is compact as a space and therefore compact as a subset of any space in which it is embedded. In particular, it is a compact subset of $\Bbb R^2$. $C$ and $D$ are open subsets of $\Bbb R^2\...


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