Is there any expression to calculate the homology groups of a quotient space? Let $B \subset A$ where $A$ is a topological space and $A/B$ the space obtained from $A$ via collapsing $B$ to a single point. 
I was wondering if there is any expression for $H_k(A/B)$ in terms of the homology groups of $A$ and $B$.
I am studying symplicial homology and I was thinking that if both spaces are triangulated spaces then it might be possible to consider the quotient of the k-simplices to get something like 
$$H_k(A/B) = H_k (A) / H_k(B) $$
But this is just an intuition and I don't think that this will hold generally. Is there any expression for the homology of the quotient?
 A: What you wrote is in general completely false. Consider $A = [0,1]$ and $B = \{0,1\}$. Then $H_1(A) = 0 = H_1(B)$, but $A/B$ is a circle and $H_1(A/B) = \mathbb{Z}$.
In general what is true is that if the inclusion $B \subset A$ is a cofibration (for example if $B$ is a sub-CW-complex of $A$), then the reduced homology group $\tilde{H}_k(A/B)$ is isomorphic to the relative homology group $H_k(A,B)$. This relative homology group fits into the long exact sequence:
$$\dots \to H_k(B) \to H_k(A) \to H_k(A,B) \to H_{k-1}(B) \to \dots$$
and in general you cannot say much more.
A: As explained in the other answer, if $(X, A)$ is a CW-pair, i.e., $X$ is a CW-complex and $A$ is a subcomplex, then one has the long exact sequence of homology groups
$$\cdots \longrightarrow \tilde{H}_{k-1}(X/A) \longrightarrow \tilde{H}_k(A) \stackrel{i_*}\longrightarrow \tilde{H}_k(X) \stackrel{q_*} \longrightarrow \tilde{H}_k(X/A) \longrightarrow \cdots$$
where $i : A \hookrightarrow X$ is the inclusion map and $q : X \to X/A$ is the quotient map. 
However, if there is a retraction $r : X \to A$, then one gets a left inverse $r_*$ of $i_*$, making the long exact sequence break up into short exact pieces 
$$0 \to \tilde{H}_k(A) \to \tilde{H}_k(X) \to \tilde{H}_k(X/A) \to 0$$
That is, $\tilde{H}_k(X/A) \cong \tilde{H}_k(X)/\tilde{H}_k(A)$. So, what you were expecting for holds whenever $(X, A)$ is a reasonably good pair, and $A$ is a retract of $X$. Note that as $r_*$ is a retract of the above sequence, you also get an isomorphism $\tilde{H}_k(X) \cong \tilde{H}_k(A) \oplus \tilde{H}_k(X/A)$ by splitting lemma. Useful corollaries of this include $\partial{D^n}$ not being a retract of $D^n$ which is equivalent to Brouwer fixed point theorem.
