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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
added 392 characters in body
Sep
6
comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
thank you! But what about my computations? Is $H_0(\mathbb{Q}) = \mathbb{Z}^\mathbb{N}$?
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 $A$
Typo corrected.
Sep
5
asked $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
Sep
5
asked $H_1(X,A) = 0 \iff H_1(A) \rightarrow H_1(X)$ surjective and $X_i$ contains no more than one path-component of $A$
Sep
5
accepted $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$
Sep
5
comment $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$
Ah no it's needed to show that the long relative homological sequence is exact!
Sep
5
comment $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$
It's needed to define $\partial_\ast$ in the chain complex $C(X)/C(A)$?
Sep
5
comment $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$
What do you mean by "it defines the complex"? If I compute $H(X,A)$ from $\dots C_n(X)/C_n(A) \xrightarrow{\partial_n} C_{n-1}(X) / C_{n-1}(A) \dots$ then I'm not sure what I need $0 \rightarrow C_0(A) \rightarrow C_0(X) \rightarrow C_0(X)/C_0(A) \dots$ for.
Sep
5
comment $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$
Yeah, I just looked at it. But then why do they mention said sequence on Wikipedia in that context?
Sep
5
comment $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$
But that's what I proved in question 15 on page 114: if $H_n(X,A) = 0$ for all $n$ then $i_\ast$ is an isomorphism for all $n$!