# Thread: Homology & path components

1. ## Homology & path components

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

What I know:

- $H_0(X)$ 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 $H_0(X,A) = 0$ iff A meets each of the path-components of X.

What I know:

- $H_0(X)$ 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

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

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

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

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

3. Thank you very much! That makes sense.