Differences in Homology/Cohomology of a CW Pair (X,A) when A is empty versus a point I am working through Hatcher, and he is proving the Kunneth Formula. At one point he claims that all that is needed is to show naturality being true in the case of the pair (X,A) having A empty implies it for A being a single point.
So my question is not about this, but rather the question in the title.
I feel like I must be misunderstanding something because I want to say their should be a natural identification of the two because I can come up with an argument showing (at least in the case of CW pairs) that they are isomorphic:
The homology groups must be identical because (X,A) is a good pair and the quotients are homeomorphic. The cohomology groups must be identical because of the Universal Coefficient Theorem.
Is there something I am missing? Does this easy isomorphism not say anything about the naturality Hatcher is talking about?
 A: Hatcher's definition of a "good pair" $(X,A)$ requires $A$ to be nonempty (see page 114).  Indeed, the theorem that the homology of $(X,A)$ is isomorphic to the reduced homology of $X/A$ is not true when $A=\emptyset$.  Following Hatcher's Proposition 2.22, we do still have an isomorphism $H_n(X,A)\cong H_n(X/A,A/A)$, but we don't have $H_n(X/A,A/A)\cong \tilde{H}_n(X/A)$ because $A/A$ is empty rather than being a single point.
(Actually, there is a way to make all the theorems still work when $A$ is empty, but it requires redefining $X/A$.  The correct definition of $X/A$ when $A$ could be empty is the quotient of $X\sqcup\{*\}$ that identifies every point of $A$ with newly adjoined point $*$.  When $A$ is nonempty this is the same as the usual quotient of $X$ that identifies all points of $A$ together, but when $A$ is empty it gives $X\sqcup\{*\}$ where the empty set $A$ has been turned into a new point.  What's really going on is that the operation sending $(X,A)$ to $X/A$ is more natural when $X$ is a pointed space, and so if $X$ is just a space you should freely turn it into a pointed space before forming $X/A$.  Or, for a better explanation, see Paul Frost's answer.)
A: The relative homology of $(X,\varnothing)$ is just that of $X$. But the relative homology of $(X,\ast)$ is the reduced homology, since it comes from the exact sequence
$$0\to C(\ast)\to C(X) \to C(X,\ast)\to 0$$
and $C(\ast)$ is the complex where $\mathbb Z$ is concentrated in degree $0$. 
A: This is not an answer, but an extended comment to Eric Wofsey's answer. Let $\mathbf T_0$ and $\mathbf T^2$ denote the categories of pointed spaces and pair of spaces, respectively. There is a canonical embedding of a categories
$$i : \mathbf T_0 \to \mathbf T^2, i(X,x_0) = (X, \{ x_0 \})$$
which simply interprets each pointed space as a pair of spaces. It has a right adjoint functor
$$Q : \mathbf T^2 \to \mathbf T_0$$
which is given by $Q(X,A) = (X/A,*)$ for $A \ne \emptyset$ and $Q(X,\emptyset) = (X^+,+)$ where $X^+ = X \cup \{ + \}$ with a point $+ \notin X$. In fact, the space $Q(X,A)$ occurs as the pushout
$\require{AMScd}$
\begin{CD}
\{ + \} @>{J}>> Q(X,A)\\
@AAA @AAA\\
A @>{j}>> X\end{CD}
where $j : A \to X$ denotes inclusion and $J(+)$ is taken as the basepoint of $Q(X,A)$.
Thus $X/\emptyset = X^+$ is not an ad hoc definition, but is a consequence of right adjointness.
For non-good pairs see my answer to relative homology is the reduced homology of the quotient.
