# Matt N.

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bio website tancast.com/wp-content/… location age member for 3 years, 6 months seen 5 hours ago profile views 5,653

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 Sep6 comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian thank you! But what about my computations? Is $H_0(\mathbb{Q}) = \mathbb{Z}^\mathbb{N}$? Sep6 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. Sep5 asked $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian Sep5 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 $A$ Sep5 accepted $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall$ path-components $X_i$ Sep5 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! Sep5 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 $C(X)/C(A)$? Sep5 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. Sep5 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? Sep5 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 $n$! Sep5 comment $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall$ path-components $X_i$ If $i_\ast$ is an isomorphism $A$ has to meet all path components of $X$. Sep5 comment $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall$ path-components $X_i$ Thank you! I think I might have found something shorter: I could use the previous question in Hatcher which states that $H_n(X,A) = 0 \iff i:A\hookrightarrow X$ induces an isomorphism.... Sep5 comment $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall$ path-components $X_i$ But they mention it on Wikipedia: en.wikipedia.org/wiki/Relative_homology. They don't explain why so I assumed they use it to compute $H_n(X,A)$. Why do they mention it there? Sep5 revised $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall$ path-components $X_i$ added 202 characters in body Sep5 accepted Determining whether there is a short exact sequence Sep5 asked $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall$ path-components $X_i$ Sep4 comment Determining whether there is a short exact sequence Thanks! I have to look at that contradiction again. Sep4 comment Determining whether there is a short exact sequence Yes, I can list the groups of order $p^{n+m}$. So you're saying what I wrote is all correct? Sep4 asked Determining whether there is a short exact sequence Sep3 comment Counter examples of cell complex OK, so in case one, there should indeed be another black dot. And in case two, the entirety of $\partial e^2$ should be mapped, not just one dot on it! Thanks for your help!