I just wanted to add something to the discussion about the utility of adding additional basepoints. It turns out this is crucial for understanding
certain aspects of embedding theory. See the bottom of this answer for an explanation.



For a map of spaces $A \to B$, let $\text{Top}(A\to B)$ be the category of spaces
which factorize this map. This has objects given by factorizations $A \to X \to B$ and 
morphisms maps $X \to X'$ which are factorization compatible in the obvious sense.
Let's consider the case of the constant map $S^0 \to \ast$. Clearly, an object of 
$\text{Top}(S^0\to \ast)$ is just a *space with a preferred pair of basepoints.*

Then **unreduced** suspension can be regarded as a functor
$$
S: \text{Top}(\emptyset \to \ast) \to \text{Top}(S^0 \to *) .
$$
That is, the functor which assigns to an unbased space its unreduced suspension,
considered as a space with two basepoints.

Now a desuspension question in this context asks given an object $X \in \text{Top}(S^0 \to *)$, is there an object $Y \in \text{Top}(\emptyset \to \ast) $ and a weak equivalence
$$
SY \simeq X ?
$$
More generally, I've gotten a lot of mileage out of the **fiberwise version** of this question. 

Given a space $B$ we can consider the unreduced fiberwise suspension of $\emptyset \to B$ as the
projection map $B \times S^0 \to B$ (here unreduced fiberwise suspension of $Y\to B$ means
the double mapping cylinder of the diagram $B \leftarrow Y \to B$, or concretely, it's
$B \times 0 \cup Y \times [0,1]\cup B \times 1$. 

**Unreduced fiberwise suspension** is then a functor
$$
S_B: \text{Top}(\emptyset \to B) \to \text{Top}(B\times S^0 \to B) ,
$$
and one can consider the problem of whether an object $X \in \text{Top}(B\times S^0 \to B)$
can be written as $S_B Y$ up to weak equivalence.

**Why I care about this problem**

This problem naturally arises in embedding theory: if $P \to N \times [0,1]$ is an embedding, where $P$ and $N$ are closed manifolds and if $W$ is the complement of $P$
in $N \times [0,1]$ then $W$ is an object of the category $\text{Top}(N\times S^0 \to N$) and a necessary obstruction to compressing $P$ as an embedding into $N$ is that $W$
should fiberwise desuspend over $N$. Furthermore, in certain instances 
the existence of fiberwise unreduced desuspension suffices to finding the compression of the embedding.  (This story is explained in detail in the paper: Poincaré duality embeddings and fiberwise homotopy theory, *Topology* **38**, 597–620 (1999).)

**Postscript**

In the fiberwise context there is a real difference between the *reduced* and *unreduced*
cases of the desuspension problem. For example, in the case of the compression problem
$P \to N \times I$ described above, the two inclusions $N \times i \to P$  for $i = 0,1$ might have distinct (fiberwise) homotopy classes. If this is the case, then there's no chance that the complement data $W$ can underly a reduced fiberwise suspension, for if it did, then
the map $N \times S^0 \to P$ would factor through $S_N N \cong N \times D^1$, giving a homotopy of the two inclusions $N \times i \to P$.

(For 
$Y \in \text{Top}(\text{id}:B \to B)$, the reduced fiberwise
suspension $\Sigma_B Y$ is given by 
$$
\Sigma_B Y = \text{colim}(B \leftarrow S_B B \to S_B Y) .
$$
This is an endo-functor of $\text{Top}(\text{id}:B \to B)$.

An even more mundane example is this: when $B = \ast$, we can consider $S^0$ with 
its two distinct basepoints. Clearly $S^0 = S\emptyset$, but $S^0$ is not, even up to weak equivalence, the reduced suspension of any based space.