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Sep
19
comment Definition of tangent space
Thanks, actually anon's read it right. I'm just generally confused at the moment by the new subject of lie algebras and groups and I didn't see that it was so obvious.
Sep
19
comment Definition of tangent space
Thanks! And thanks for the book recommendation. As for the bonus: I'm still struggling with definitions so I'm not quite ready to answer that.
Sep
17
comment Exact meaning of homology
I picture a one dimensional hole as something you can not get $S^1$ passed. I'm not sure it's right but in a cell complex such as the torus I can consider the one skeleton and then I see that it contains $2$ one dimensional holes. Now going to look at the other questions you gave me... thank you!
Sep
8
comment $X$ $n$-connected $\iff $ any continuous map $f:K \rightarrow X$ is null-homotopic
@Shaun Ault: But $X$ is not necessarily a CW complex. It's just a topological space.
Sep
8
comment $X$ $n$-connected $\iff $ any continuous map $f:K \rightarrow X$ is null-homotopic
@Michael: I thought that too, several times even. But then I always thought I'm just not advanced enough yet...
Sep
7
comment $X$ a CW complex is contractible if it's the union of an increasing sequence
I take $x_0$ and the inclusion $i: \{ x_0 \} \hookrightarrow X$. $i_\ast$ is a map from zero to zero and is therefore an isomorphism and then by Whitehead a homotopy equivalence?
Sep
7
comment $X$ a CW complex is contractible if it's the union of an increasing sequence
Thank you! I don't see how this follows from the Whitehead theorem though...
Sep
7
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
So a basis for $\operatorname{ker}{g}$ is $\{ [q]_\mathbb{Q} : q \in \mathbb{Q} - \ast \}$?
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
Oh! I just noticed that you edited your answer! Thank you!
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
But I don't need to have $g$ explicitly. All I need is its kernel and I can compute that without knowing $g$. Am I missing anything?
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
I updated it, that should be the rationals minus a point.
Sep
6
comment $H_1(X,A) = 0 \iff H_1(A) \rightarrow H_1(X)$ surjective and $X_i$ contains no more than one path-component of $A$
Yes, of course! I didn't know I could do that, thanks!
Sep
6
comment Homology groups of unit square with parts removed — revisited
So in that case, it's only different to my other solution in the case $n=1$...
Sep
6
comment Homology groups of unit square with parts removed — revisited
If I take $X$ to be the boundary plus the line in the middle at $\frac{1}{2}$ this is homotopy equivalent to a wedge of two copies of $S^1$, so $H_n(X) = H_n(S^1) \oplus H_n(S^1) = 0$ for $n > 1$ and $\mathbb{Z} \oplus \mathbb{Z}$ for $n=1$ and $\mathbb{Z}$ for $n=0$.
Sep
6
comment Homology groups of unit square with parts removed — revisited
Thanks! That was actually just a typo, thanks for pointing it out!
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
OK, I have a better argument. Continuous functions map connected spaces to connected spaces. $[0,1]$ is connected, $\mathbb{Q}$ is not, therefore there cannot be a continuous function $[0,1] \rightarrow \mathbb{Q}$?
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
There are so many words I don't know the definition of! But so $H_0(\mathbb{Q}) = \oplus_{q_i} \mathbb{Z}$? Would that be correct if I wrote it like that?
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
So what's the difference between the direct sum and the direct product in this case? I thought they were the same!
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
If it maps to more than one point, say to two, $q_1, q_2$, then for every delta their distance will be greater equals $|q_1 - q_2|$. So pick epsilon smaller than this distance and you cannot find a delta such that continuity holds...
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
Every continuous function has to be constant, equal to $q_i$?