suspension of the torus The suspension of $S^n$ is $S^{n+1}$. What about the suspension of $T^n$? Is it just $S^2\times T^{n-1}$? I'm having a hard time visualizing even the first non-trivial case $\Sigma T^2$.
 A: The cartesian product of two manifolds is a manifold, so $S^2 \times T^{n-1}$ is a manifold. The suspension of the torus, however, is not. In fact, the suspension of a manifold is almost never a manifold – the only counterexample is the sphere (proof: a it's a standard exercise that $\tilde{H_n}(X)=\tilde{H}_{n+1}(\Sigma X)$ where $\Sigma X$ is the suspension. See Hatcher, section 2.1, Exercise 20. Now use Poincaré duality.) 
A good way to imagine the suspension $\Sigma T^2$ is to imagine $T^2$ as a flat square, where you know the edges are pairwise identified but you don't actually fold the square to affect the identification. The suspension is now obvious: build two pyramids with the square as their base, one above and one below, to obtain a kind of octahedron. 
Now identify the edges. Of course, you need to identify them all the way up and down.

(Image source)
The tips of the suspension are singularities. They don't have neighborhoods homeomorphic to Euclidean space; their links are tori, not spheres. 
A: It is probably better to work this out algebraically in the homotopy category of pointed space because there is a formula for this.  We use $\cong$ to denote homotopy equivalence.
Let $X$ be a topological space with a base $x$.  Then the suspension operation $S(X)$ turns $\{x\} \times I$ into an interval.  In order to obtain a space with a base point, we contract this $\{x\} \times I$ into a point and obtain $\Sigma(X)$, the reduced suspension of $X$.  Suppose $X,Y$ are spaces with base points $x,y$, respectively.  Then we can join $X, Y$ at the base points and obtain a new pointed space $X \vee Y$.  Then we have a general formula:
$\Sigma(X \times Y) \cong \Sigma(X) \vee \Sigma(Y) \vee \Sigma(X \wedge Y)$.
The $\wedge$ is the smash product which is a bit complicated, but we have the relation:
$X \wedge S^1 \cong \Sigma(X).$
In our particular case we can get rid of it using induction.
To do this, substituting $T^1 = S^1$ for $X$ and $Y$, we get
$\Sigma(T^2) = \Sigma(T \times T) \cong \Sigma(S^1) \vee \Sigma(S^1) \vee \Sigma(S^1 \wedge S^1) \cong S^2 \vee S^2 \vee S^3.$
These and together with the fact that
$\Sigma(S^n) \cong S^{n+1}$ will give you $\Sigma(T^n)$ by induction on $n$.
