17,750 reputation
53497
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visits member for 3 years, 6 months
seen 10 hours ago

Aug
26
asked Proof that inclusion in disconnected space cannot be null-homotopic
Aug
26
asked Inclusion of a neighbourhood is nullhomotopic
Aug
25
accepted Meaning of $( \alpha_i = ((0,\dots, a_n, \dots, 0)) $ converges in $\mathbb{R}^\infty$
Aug
25
comment Cellular homology of projective space $\mathbb{R}P^n$
Oooh! So half of it is glued around one direction and the second half of it is glued around the opposite direction so that it becomes the $0$ map?
Aug
25
comment Homology relative to a point
But the augmentation map is at the end of the chain, like so: $0 \rightarrow C(\ast) \xrightarrow{i_\ast} C_0(X) \xrightarrow{j_\ast} C_0(X, \ast) \xrightarrow{\varepsilon} \mathbb{Z} \rightarrow 0$ so how can it be an inverse for $i_\ast$?
Aug
25
comment Cellular homology of projective space $\mathbb{R}P^n$
Yes I did that already, it's multiplication by $2$!
Aug
25
asked Cellular homology of projective space $\mathbb{R}P^n$
Aug
25
comment Homology relative to a point
thanks! I understand that $0 \rightarrow C(\ast) \xrightarrow{i_\ast} C_0(X) \xrightarrow{j_\ast} C_0(X, \ast) \rightarrow 0$ splits because $i_\ast$ has an inverse (the constant map is a left inverse). From that it follows that $C_0(X) \cong C_0(X, \ast) \oplus C_0(\ast )$. How did you get $\tilde{C_0} (X) \cong C_0(X, \ast) \oplus \mathbb{Z} \cong C_0(X, \ast) \oplus C_0(\ast)$ from that?
Aug
23
comment Homology relative to a point
I'm not sure where to go from here: $0 \rightarrow C_0(\ast) \xrightarrow{i} C_0(X) \xrightarrow{j} C_0(X, \ast ) \xrightarrow{\varepsilon} \mathbb{Z} \xrightarrow{0} 0$.
Aug
22
comment Homology relative to a point
The homology groups that come from the extended chain complex where one inserts $\mathbb{Z}$ at the end of the sequence. (the definition in Hatcher)
Aug
22
asked Homology relative to a point
Aug
22
comment Klein bottle homology by Mayer Vietoris
Thank you! ....
Aug
22
accepted Klein bottle homology by Mayer Vietoris
Aug
21
comment Klein bottle homology by Mayer Vietoris
Thank you! But what is $2 \mathbb{Z} (1,1)$? First I thought it's the free abelian group over the basis consisting of one element, $(1,1)$ but then that would be $\mathbb{Z}$ and I need it to be $\mathbb{Z} \oplus 2\mathbb{Z}$ but I don't see how $2 \mathbb{Z} (1,1) = \mathbb{Z} \oplus 2\mathbb{Z}$...
Aug
21
asked Klein bottle homology by Mayer Vietoris
Aug
21
accepted Follow-up on $H_n(\mathbb{R}^3 - S^1)$
Aug
21
revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$
edited body
Aug
21
comment Follow-up on $H_n(\mathbb{R}^3 - S^1)$
@Dylan: thank you so much!!
Aug
21
revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$
edited body
Aug
20
accepted Excision in homology: $H(D^2, S^1)$