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

0
votes
Let us denote $G=\pi_1(X,x_0)$ and $H=p_*(\pi_1(\bar{X},\bar{x}_0)$. The first statement, about the cyclicity of the group $\text{deck}(X,p)$ of deck transformations, follows when you observe that $\ …
answered Jan 13 '15 by Espen Nielsen
3
votes
0answers
I have to show that if $X$ is a CW-complex, $A$ is a subcomplex and $i:A\hookrightarrow X$ is a cellular inclusion, then $i$ is a cofibration. My attempt is as follows. I think my proof, which first p …
asked Jan 24 '13 by Espen Nielsen
3
votes
We talk about lifting a map $f$ through another map $\rho$. The map $\rho$ is therefore a part of the data, together with the spaces $X,\tilde{X}$ and the map $f$.
answered May 5 '14 by Espen Nielsen
0
votes
Given a smooth curve $\gamma:I\rightarrow \mathbb{R}^2$, You can define the Gauss map $n:I\rightarrow S^1\subset \mathbb{R}^2$ giving you a unit vector normal to your curve at every point in a smooth …
answered Nov 2 '12 by Espen Nielsen
1
vote
I will show you that whevener you have chosen a homotopy $H$, you can construct a new homotopy which is constant on $S^{n-1}$. First, we label points on $D^n$ as $(r,\theta)$, treating $r$ as a cont …
answered Jan 14 '15 by Espen Nielsen
7
votes
2answers
I think I found a proof of Brouwer's fixed point theorem which is much simpler than any of the proofs in my books. One part is standard: Suppose there is an $f:D^n \rightarrow D^n$ with no fixed poin …
asked Nov 30 '12 by Espen Nielsen
15
votes
3answers
It is easy to show that for any topological space $X$, the cone $CX$ is contractible. I am interested in the converse. If $Y$ is a contractible space, is $Y$ homeomorphic to $CX$ for some topological …
asked Jan 16 '13 by Espen Nielsen
2
votes
If you have two continuous maps $f:Z\rightarrow X$ and $g:Z\rightarrow Y$, the adjunction space (or pushout, as it is usually called), is the space $W=X\coprod_ZY=(X\coprod Y)/\sim$, where $\sim$ is t …
answered Nov 2 '14 by Espen Nielsen
13
votes
1answer
So far, any source I consult will gladly talk about cobordism classes of closed (compact and without boundary) oriented manifolds, but I have yet to see an example of a pair of manifolds which are not …
asked Jan 22 '13 by Espen Nielsen
0
votes
For part 1: The part where you (between lines) write $H_{n-1} \circ \partial '_n \circ i_n = -h_{n-2}\circ \partial^X_{n-1}$ is the error. Remember that $\partial '(0,0,z)=(-z,z,-\partial^X z)$. Then …
answered Feb 4 '15 by Espen Nielsen
11
votes
For a quick introduction, you can read this AMS survey. What is … Persistent Homology? by Shmuel Weinberger A basic notion in persistent homology is a barcode. The following article gives an intro …
answered Oct 15 '15 by Espen Nielsen