Can we change $\tilde{H}$ to $H$ in the following theorem? There is this theorem about the sequence of homology group of a space, a subspace and its quotient space.

My question is why the reduced homology group $\tilde{H}$ is used instead of the usual $H$? If we change all the $\tilde{H}$'s to $H$'s will the theorem still be correct?
All I understand about the difference between the reduced homology group and the usual one is that $\tilde{H}_0(X)=H_0(X)\oplus\mathbb{Z}$ but $\tilde{H}_i(X)=H_i(X)$ for $i>0$, right? Is there any other specific purposes for introducing the reduced homology?
 A: What is true in general is the following: if $X$ is a space and $A\subseteq X$ is a subspace, then there is a long exact sequence $$\dots\to h_n(A)\to h_n(X)\to h_n(X,A)\to h_{n-1}(A)\to\cdots$$
where $h_n$ is either $H_n$ or $\tilde{H}_n$ everywhere (but you make the same choice in the entire sequence).  Indeed, this is just the long exact sequence on homology associated to the short exact sequence of chain complexes $$0\to c_\bullet(A)\to c_\bullet(X)\to c_\bullet(X,A)\to 0,$$ where $c_\bullet$ means either the singular chain complex or the augmented singular chain complex.
Now, what does this have to do with the theorem you're asking about?  Well, under certain hypotheses (such as the hypotheses of the theorem), there is a natural isomorphism $h_n(X,A)\cong \tilde{H}_n(X/A)$ (where again $h_n$ could be either $H_n$ or $\tilde{H}_n$).  Taking $h_n=\tilde{H}_n$, this gives your theorem.  But you could also take $h_n=H_n$, and then you would get a long exact sequence $$\dots\to H_n(A)\to H_n(X)\to \tilde{H}_n(X/A)\to H_{n-1}(A)\to\cdots.$$
However, you cannot get a long exact sequence using unreduced $H_n(X/A)$,  For instance, suppose that $X=A$ is just a point, so $X/A$ is also a point.  Then the end of the sequence $$H_0(A)\to H_0(X)\to H_0(X/A)\to 0$$ becomes $$\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}\to 0$$ where the maps are the identity.  This is not exact at the second term.
