Relative homology and path connected space I want to prove that if $X$ is a path connected space and if $Y$ is nonempty then $$H_0(X,Y)\simeq 0$$
it is sayed that we have this chain: $H_0(Y)\rightarrow H_0(X)\rightarrow H_0(X,Y)\rightarrow 0$ and then  is surjetive and then we obtain the result ! 
but i dont understand why $H_0(Y)\rightarrow H_0(X)$ is surjective (i know that $H_0(X)\simeq \mathbb{Z}$) but i don't understand how we can obtain the result using this surjetivity ?
please help me thank you. 
 A: I think it's useful to know the following fact:  $i_*:H_0(A)\to H_0(X)$, induced by the inclusion $i:A\hookrightarrow X$ of subspace $A$, is surjective if and only if $A$ meets each path-component of $X$.
The group $H_0(X)$ is free abelian and a basis is given by the classes $[x_i]$ where $X_i,\ i\in I$, is the set of path components of $X$ and $x_i\in X_i$.
If $A$ meets each path component of $X$, then for each $x_i$ there is an $a_i\in A$ such that $i(a_i)\in X_i$. Since there is a path from $i(a_i)$ to $x_i$, their difference $x_i-i(a_i)$ is a boundary, thus $[x_i]=i_*([a_i])$ and $i_*$ is onto.
Conversely, if $i_*$ is onto, then for each $i\in I$ there is an $a_i\in A$ with $[i(a_i)]=[x_i]$. But this means that $x_i-a_i$ is a boundary. We know that the boundary of a $1$-chain is the sum of the boundaries on the individual path components, where each of these boundaries is a formal sum of points such that the sum of the coefficients $0$. That means that if $x_i$ appears in the boundary on one component, then $a_i$ must cancel it out, so it must lie in the same path component.
Now, $H_0(X,A)$ is zero iff $H_0(X)→H_0(X,A)$ is the zero map iff $H_0(A)→H_0(X)$ is surjective.
