# Matt N.

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 Sep25 comment $C_c(X)$ dense in $L_1(X)$ Thank you! About the measurable function with compact support: the step function $s_n$ has finite support and therefore compact support. This is what I meant. Sep24 comment $C_c(X)$ dense in $L_1(X)$ Did you use the Tietze theorem anywhere? Sep24 comment $C_c(X)$ dense in $L_1(X)$ Thanks! Where do you get the increasing sequence from? $\sigma$-finite doesn't give you that: en.wikipedia.org/wiki/%CE%A3-finite_measure Sep19 comment Lie algebra of $GL_n(\mathbb{C})$ I know nothing. I started to read about Lie groups yesterday. I read that the exponential map maps elements in the Lie algebra to elements in the Lie group. I wonder if that is useful. Probably not because the exponential map is not surjective, I suppose.... Sep19 comment Definition of tangent space Thanks, actually anon's read it right. I'm just generally confused at the moment by the new subject of lie algebras and groups and I didn't see that it was so obvious. Sep19 comment Definition of tangent space Thanks! And thanks for the book recommendation. As for the bonus: I'm still struggling with definitions so I'm not quite ready to answer that. Sep17 comment Exact meaning of homology I picture a one dimensional hole as something you can not get $S^1$ passed. I'm not sure it's right but in a cell complex such as the torus I can consider the one skeleton and then I see that it contains $2$ one dimensional holes. Now going to look at the other questions you gave me... thank you! Sep8 comment $X$ $n$-connected $\iff$ any continuous map $f:K \rightarrow X$ is null-homotopic @Shaun Ault: But $X$ is not necessarily a CW complex. It's just a topological space. Sep8 comment $X$ $n$-connected $\iff$ any continuous map $f:K \rightarrow X$ is null-homotopic @Michael: I thought that too, several times even. But then I always thought I'm just not advanced enough yet... Sep7 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? Sep7 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... Sep7 comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian So a basis for $\operatorname{ker}{g}$ is $\{ [q]_\mathbb{Q} : q \in \mathbb{Q} - \ast \}$? Sep6 comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian Oh! I just noticed that you edited your answer! Thank you! Sep6 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? Sep6 comment $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian I updated it, that should be the rationals minus a point. Sep6 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! Sep6 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 $n=1$... Sep6 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 $n=0$. Sep6 comment Homology groups of unit square with parts removed — revisited Thanks! That was actually just a typo, thanks for pointing it out! Sep6 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 $[0,1] \rightarrow \mathbb{Q}$?