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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
revised $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
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Sep
6
revised $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
added 534 characters in body
Sep
6
revised $H_1(X,A) = 0 \iff H_1(A) \rightarrow H_1(X)$ surjective and $X_i$ contains no more than one path-component of $A$
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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
revised Homology groups of unit square with parts removed — revisited
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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
revised Homology groups of unit square with parts removed — revisited
Typo correction.
Sep
6
asked Homology groups of unit square with parts removed — revisited
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 $q_i$?
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
Because every path-component (i.e. point $q_i$) generates a homology group that is isomorphic to $\mathbb{Z}$. Then I do $H_0(\mathbb{Q}) = \oplus H_0(q_i) = \mathbb{Z}^\mathbb{Q} = \mathbb{Z}^\mathbb{N}$. Where is my mistake?
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
Well, that's what I'm very insecure about. Is it or is it not, $\mathbb{q_i}$? Every point in $\mathbb{Q}$ is a path-component of it I think. But then I'd have $\mathbb{Z}^\mathbb{N}$ and I'm not sure about that...
Sep
6
revised $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
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