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Questions tagged [algebraic-topology]

Questions about algebraic methods and invariants to study and classify topological spaces: homotopy groups, (co)-homology groups, fundamental groups, covering spaces, and beyond.

Algebraic topology is a mathematical subject that associates algebraic objects to topological spaces which are invariant under homeomorphism or homotopy equivalence. Often, these associations are functorial so that continuous maps induce morphisms between appropriate algebraic objects.

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21356 questions
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Connecting two Hawaiian earings by a line segment - is the inclusion a cofibration?

Suppose that I have two Hawaiian earrings, and i connect them by a line segment (through the wedge points). I would like to inquire whether the inclusion of the line segment is a cofibration. My confusion is: On one hand, it seems to me that the…
902
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Induced map between zeroth homology groups an isomorphism?

Suppose that $n,m$ are integers such that $n > m$. I know that the singular homology groups $H_0(\Bbb{R}P^n;\Bbb{Z})$ and $H_0(\Bbb{R}P^m;\Bbb{Z})$ are both isomorphic to $\Bbb{Z}$. Now suppose I look now at homology with $\Bbb{Z}/2\Bbb{Z}$…
user38268
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Given G an abelian group finitely generated, and looking for a simplicial complex K such that $H_{1}(K,\mathbb{Z})=G$

the problem statement is as follows: Given G an abelian group, I have to show that there exist a simplicial complex, K (finite), such that the $H_{0}(K,\mathbb{Z})=\mathbb{Z}$ and $H_{1}(K,\mathbb{Z})=G$. The first thing that came to my mind was use…
Monkey
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Inclusion induces epimorphism in homology

I am given $A,B\subseteq X$ such that $Int A\cup Int B=X$. I need to prove that inclusion $(X,A\cap B)\hookrightarrow (X,A)$ induces epimporhism in realtive homology groups. Is this a right argument: From the naturality of long exact homology…
Madara Uchiha
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Example of two spaces with same cohomology ring but different homotopy group

I want to find example of two spaces with same cohomology ring but different homotopy group. I know spaces with same homotopy groups but different homology groups and vice versa. But currently I have not found example of spaces with same cohomology…
user330928
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Some specific property of 3-dimensional closed orientable manifold

I am doing a problem which ask to show the following: Assume we have a 3-dimensional closed orientable manifold M with fundamental group $Z_p$, then M is not homeomorphic to any subspace of $Y$ of $S^4$ such that the complement space of $Y$ in $S^4$…
user330928
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Involution map on $D^2$ with identity map on boundary is identity

I am trying to prove the claim as the title. Here by involution map I mean a map after composing itself becomes identity. So if such map restricts to the boundary of the disk, it turns out that this map must be identity on the whole disk. I only…
user330928
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$R^3$ with figure 8 removed?

I am thinking whether there is a simple space equivalent to $R^3$ with a figure 8 removed from xy-plane. And what about wedge sum of n circles is removed? For only one circle, the complement space actually is homotopic equivalent to a circle wedge…
user330928
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$X=S^1\times D^2/S_1\times S_1$ is homotopy equivalent to $S^2\vee S^3$

I'm looking for a proof that $X=S^1\times D^2/S_1\times S_1$ is homotopy equivalent to $S^2\vee S^3$. I can't think of any rigorous proof and I'm not even sure if I can see why this is true. Can you help me? (If this helps, I've proved $S^n/S^k$ is…
Jules
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Why is the "house with two rooms" is a 2-dimensional subspace of $\mathbb{R}^3$?

We know that a plane is a two-dimensional subspace of $\mathbb{R}^3$. But in Hatcher's Algebraic Topology, the house with two rooms is a $2$-dimensional subspace of $\mathbb{R}^3$. Why is that? Isn't it supposed to be a $3$-dimensional subspace of…
user338393
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Definition of cone over $X$

The cone over $X$, as we know, is defined as $CX=X\times I/\{x\}\times I$ where the geometric picture is clear to me. But I saw another definition where $CX$ is defined as $X\times I/\{x\}\times I\cup X \times \{1\}$ which is not visually very…
SAUVIK
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Why does the only non-orientable surface embeddable in $\mathbb{R}^3$ have boundary?

As far as my understanding goes, if one has a closed (compact) surface and "cuts out" a disk, then the resulting object has a non-empty/non-trivial boundary. It can be shown that any non-orientable surface without boundary cannot be embedded into…
Chill2Macht
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Intersection pairing in cohomology and Poincare duality

This question is similar to but not exactly the same as the old question (the issue of torsion). In Exercise 11.8 of Dan Freed's notes he claims that the following pairing being nondegenerate is equivalent to Poincare duality $$\bar{I}_M:…
PhysicsMath
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Possible rank of closed orientable manifolds

List all $i$ for which there is a closed orientable $6$-manifold $ M$ with $H_i(M) = \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$. So, obviously this has to do with Poincare Duality. We can't have $i=0$ or $i=6$ by definition of a manifold. So…
Ng-ettingit
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The Thom spectrum of the stable framing tangential structure is the sphere spectrum

where a stable framing is a stable tangential structure $\mathcal{X} = EO \to BO$ (ref. Dan Freed's notes Exercise 9.50). This is Exercise 10.32 in Dan Freed's notes and I have no idea to get started with the proof. Could somebody sketch the main…
PhysicsMath
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