# Homology maps

1. Jul 14, 2007

### bham10246

Hi, I'm working on some homology problems but I need help figuring out the induced map from a given map, say $f:X\rightarrow Y$.

For example, compute $H_* (\mathbb{R}, \mathbb{R}^n - p)$ where $p \in \mathbb{R}^n$.

So for $n=1$, we have the long exact sequence
$0 \rightarrow H_1(\mathbb{R}^n-p)=0 \rightarrow H_1(\mathbb{R}^n)=0 \rightarrow H_1(\mathbb{R}^n, \mathbb{R}^n-p)$
$\rightarrow H_0(\mathbb{R}^n-p)=\mathbb{Z}^2 \rightarrow H_0(\mathbb{R}^n)=\mathbb{Z} \rightarrow H_0(\mathbb{R}^n, \mathbb{R}^n-p)\rightarrow 0.$

So I got $H_1(\mathbb{R}^n, \mathbb{R}^n-p)=\mathbb{Z}$ but what is $H_0(\mathbb{R}^n, \mathbb{R}^n-p)$? Is $H_0(\mathbb{R}^n, \mathbb{R}^n-p)=\mathbb{Z}$ or $0$? I first thought it was $\mathbb{Z}$ because it's path connected, but I'm not so sure anymore. It's because $H_0 (\mathbb{R}^n-p)=\mathbb{Z}^2$. So since $(\mathbb{R}^n-p) \rightarrow \mathbb{R}^n$ is an inclusion map, what is the induced map $H_0 (\mathbb{R}^n-p) \rightarrow H_0(\mathbb{R}^n)$? Is it injective or surjective? If I have this one key information, then I'm sure I can deduce the relative homology groups.

Now for $n = 2$, we have
$0 \rightarrow H_2(\mathbb{R}^n-p)=0 \rightarrow H_2(\mathbb{R}^n)=0 \rightarrow H_2(\mathbb{R}^n, \mathbb{R}^n-p)$
$\rightarrow H_1(\mathbb{R}^n-p)=\mathbb{Z} \rightarrow H_1(\mathbb{R}^n)=0 \rightarrow H_1(\mathbb{R}^n, \mathbb{R}^n-p)$
$\rightarrow H_0(\mathbb{R}^n-p)=\mathbb{Z} \rightarrow H_0(\mathbb{R}^n)=\mathbb{Z} \rightarrow H_0(\mathbb{R}^n, \mathbb{R}^n-p)\rightarrow 0.$.

I know $H_2(\mathbb{R}^2, \mathbb{R}^2-p)=\mathbb{Z}$ but are $H_1(\mathbb{R}^2, \mathbb{R}^2-p)$ and $H_0(\mathbb{R}^2, \mathbb{R}^2-p)$ isomorphic to the integers as well? If so, why?

Again, I think if I understand the following: if $f:A\rightarrow X$ is an inclusion map, is it always true that the induced map must be surjective? Can it be injective as well?

Thank you so much for your help!

Last edited: Jul 14, 2007
2. Jul 14, 2007

### mathwonk

there seem to be a lot of errors here.

3. Jul 14, 2007

### bham10246

Hi, sorry. I copied and pasted some of the math fonts, and I see some mistakes now. I'll fix them.

Okay, the mistakes I saw have been fixed. But there are more mistakes I'm sure.

Last edited: Jul 14, 2007
4. Jul 15, 2007

### matt grime

I would really go and check your facts, bham. There still seem to be a lot of errors.

5. Jul 15, 2007

### mathwonk

please remind me abiout relative homology. is it usually the homology of the space obtained by collapsing the subspace to a point? if so, would that mean that this one i the same as the homology of the one point compactification of R^n? i.e. of an n sphere?

6. Jul 15, 2007

### bham10246

Okay, under the fundamental group section on page 50 in Hatcher, I found the following Proposition: if $A \rightarrow X$ is an inclusion map, then the induced map $i_*: \pi _1(A) \rightarrow \pi _1 (X)$ is surjective. The problem was that I couldn't find an analoguous version of the Proposition for homology groups. But if I suppose that this Proposition holds true with homology groups, then for n=1, let

$\psi_1: H_1(\mathbb{R}^1)\rightarrow H_1(\mathbb{R}^1, \mathbb{R}^1-p)$,
$\delta_1: H_1(\mathbb{R}^1, \mathbb{R}^1-p)\rightarrow H_0(\mathbb{R}^1-p)$,
$\phi_0: H_0(\mathbb{R}^1-p)\rightarrow H_0(\mathbb{R}^1)$,
and
$\psi_0: H_0(\mathbb{R}^1)\rightarrow H_0(\mathbb{R}^1,\mathbb{R}^1-p )$.

Then since n=1, $H_1(\mathbb{R}^1)=0$ because the real line is contractible.
$H_0(\mathbb{R}^1-p)=\mathbb{Z}^2$ because the two intervals $(-\infty, p) \cup (p,\infty)$ contracts to two points, and the zeroth homology can be thought of as a union of distinct path components.
Finally, $H_0(\mathbb{R}^1)=\mathbb{Z}$ because the real line contracts to a point and one point is path connected.

Because $\psi_1$ is the zero map, $\delta_1$ is injective. So $H_1(\mathbb{R}^1, \mathbb{R}^1-p)$ is isomorphic to $im\delta_1 = \ker \phi_0$.

Since $\mathbb{R}^1-p \rightarrow \mathbb{R}^1$ is an inclusion map, the induced map $\phi_0$ is surjective (I will assume that the homotopy version of the above Proposition holds true with homology groups). So $im\phi_0=\mathbb{Z}$. So $\frac{H_0(\mathbb{R}^1-p)}{\ker\phi_0}=\frac{\mathbb{Z}^2}{\ker\phi_0}$ is isomorphic to $im\phi_0=\mathbb{Z}$. So $\ker\phi_0=\mathbb{Z}$.

Thus $H_1(\mathbb{R}^1, \mathbb{R}^n-p)=\mathbb{Z}$.

Next, $\frac{H_0(\mathbb{R}^1)}{\ker\psi_0} =\frac{\mathbb{Z}}{im\phi_0}=\frac{\mathbb{Z}}{\mathbb{Z}}=0$ is isomorphic to $im \psi_0 = H_0(\mathbb{R}^1,\mathbb{R}^1-p)$. So $H_0(\mathbb{R}^1,\mathbb{R}^1-p)=0$.

As for higer dimensions, I can deduce from the base case.

So matt grime, am I now correct or am I still incorrect? Do theorems for the homotopy theory continue to hold true in homology theory (the only difference is that in homology theory, all groups have been abelianized)?

And mathwonk, yes, it is the homology of the space obtained by collapsing the subspace to a point. But I don't think it's the same as the homology of the one-point compactification of R^n because in that case, we have $H_0(S^n)=\mathbb{Z}$, $H_n(S^n)=\mathbb{Z}$ and $H_j(S^n)=0$ for all j different from 0 and n. Matt grime, can you help us out?

Last edited: Jul 15, 2007
7. Jul 16, 2007

### matt grime

that seems better - what you wrote in the first post confused me since, I can now see, you say n=1, but then leave n=n in the working. Please, if you're going to let n=1 use notation like R^1, not R^n to spare me the head scratching.

8. Jul 16, 2007

### bham10246

haha... thank you so much for your reply and confirming whether I am right or wrong! It took me 2 hours on Sunday to type everything! Wow... combining LaTeX and regular typing is hard!!