Relative homology can be represented by a homology of quotient space? $H_*(X,Y) = H_*(X/Y)$? Let $X,Y$ be topological spaces, and $Y \subset X$.
Let $X/Y$ be the quotient space by the relation $x \sim y$ iff $x=y$ or $x,y \in Y$.
Then $H_*(X,Y) = H_*(X/Y)$ is true or not?
Similarly $\pi_*(X,Y) = \pi_*(X/Y)$ is true?
 A: As pointed out in the comments it is not true that $\pi_*(X,Y)$ and $\pi_*(X/Y)$ are isomorphic. Take $(D^n,S^{n-1})$. Then $\pi_k(D^n,S^{n-1})\cong\pi_{k-1}S^{n-1}$, since $\pi_lD^n=0$ for all $l$. On the other hand $D^n/S^{n-1}\cong S^n$. Take $n=2$. Then
$$\pi_3(D^2,S^1)\cong \pi_2S^1=0$$
whilst
$$\pi_3S^2\cong\mathbb{Z}.$$
Things are better with homology, although there are still restrictions. If the inclusion $Y\subseteq X$ is a cofibration, then the quotient map induces an isomorphism
$$H_n(X,Y)\cong H_n(X/Y,\ast)=\widetilde H_n(X/Y)$$
for all $n\geq0$. It is sufficient that $Y\subset X$ is a cofibration, although not strictly necessary. For instance let $X\subseteq\mathbb{R}^2$ be the comb space and $Y=\{(0,1)\}$. Then $Y\subseteq X$ is not a cofibration, although $H_*(X,Y)$ and $H_*(X/Y)$ vanish identically.
On the other hand, the isomorhism in question can fail horrendously. Let
$$X=\{1/n\mid n\in\mathbb{N}\}\cup \{0\}\subseteq\mathbb{R}$$
and equip it with the base point $0$. Write $\Sigma X$ for the unreduced suspension of $X$ and $\widetilde\Sigma X$ for the reduced suspension of $X$ obtained as the quotient space $\widetilde\Sigma X=\Sigma X/\{0\}\times I$. Note that $\widetilde\Sigma X$ is the infamous Hawaiian Earring, aka the shrinking wedge of circles.
Then $\Sigma X$ is path connected and it is known that $\pi_1(\Sigma X)$ is free on a countable infinite set of generatators. Thus $H_1(\Sigma X)$  is free abelian on countably many generators. Since $\{0\}\times I$ is contractible, the long exact sequence for the pair $(\Sigma X,\{0\}\times I)$ gives rise to an isomorphism
$$H_1(\Sigma X)\cong H_1(\Sigma X,\{0\}\times I).$$
On the other hand, the first singular homology of the Hawaiian Earing $H_1(\widetilde\Sigma X)$ contains an uncountable number of copies of $\mathbb{Q}$.
