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Sep
7
accepted $X$ a CW complex is contractible if it's the union of an increasing sequence
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
asked $X$ a CW complex is contractible if it's the union of an increasing sequence
Sep
7
asked $X$ $n$-connected $\iff $ any continuous map $f:K \rightarrow X$ is null-homotopic
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
revised $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
added 69 characters in body
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 $A$
added 85 characters in body
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
added 53 characters in body
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 $[0,1] \rightarrow \mathbb{Q}$?