# Thread: Homology & path components

1. ## Homology & path components

I have to show that $\displaystyle H_0(X,A) = 0$ iff A meets each of the path-components of X.

What I know:

- $\displaystyle H_0(X)$ is the direct sum of n copies of Z where n is the # of path-components in X
- A will have at least as many path-components as X

What I don't know:

- How to use relative Homology to prove the result

Help!

2. Originally Posted by harbottle
I have to show that $\displaystyle H_0(X,A) = 0$ iff A meets each of the path-components of X.

What I know:

- $\displaystyle H_0(X)$ is the direct sum of n copies of Z where n is the # of path-components in X
- A will have at least as many path-components as X

What I don't know:

- How to use relative Homology to prove the result

Help!
The long exact sequence has the following form

$\displaystyle ...H_0(A) \rightarrow H_0(X) \rightarrow H_0(X,A) \rightarrow 0$ (Hatcher p 115).

Since A has at least as many path-components as X, $\displaystyle H_0(A) \rightarrow H_0(X)$ is surjective (They are both free abelian groups and H_0(A) has the same or higher rank than H_0(X)).

Let $\displaystyle M:H_0(A) \rightarrow H_0(X)$ and $\displaystyle N:H_0(X) \rightarrow H_0(X, A)$. By the definition of the long exact sequence, $\displaystyle \text{Im M = Ker N}$.

Since $\displaystyle H_0(A) \rightarrow H_0(X)$ is surjective, it forces N is a zero map and $\displaystyle H_0(X,A)=0$ (The kernel of N should be the whole image of M, which is H_0(X)) .

3. Thank you very much! That makes sense.