Showing that two spaces are homeomorphic I was trying to show that a torus is homeomorphic to $S^1 \times  S^1$ ,  I tried to work with the fundamental group of both, which are equal, but that doesn't imply they're homeomorphic, (at least i haven't took it yet )so I'm stuck I can't think of anything else , 
I can see geometrically how the circles form the torus , but I can't write it down .  Can anyone help me please ?
 A: Hint:
The standard parametrization of a torus in $\mathbb{R}^3$ is 
$$  (\theta,\phi)\mapsto ((r+a\cos\theta) \cos\phi,( r+a\cos\theta )\sin\phi, a \sin\theta)$$
This can be rewritten as
$$ (\theta,\phi)\mapsto (r\cos\phi,r\sin\phi,0) + \big[a\cos\theta(\cos\phi,\sin\phi,0) + a\sin\theta(0,0,1)\big]. $$
Stare at this hard, and remember that $\phi$ and $\theta$ run between $0$ and $2\pi$. Do you see how this answers your question?
(You can always compute the Jacobian, show that it's everywhere invertible, and observe that the function is injective on $[0,2\pi]\times[0,2\pi]$, but that's much less satisfying than seeing exactly how this parametrizes a copy of $S^1\times S^1$.)
A: The torus can be defined as a square with opposite edges identified.  It can be realized geometrically as the image of the map of the plane into 4 space defined by
$$(x,y) \rightarrow (\sin(x), \cos(x), \sin(y), \cos(y))$$
This map is 2 pi periodic in both x and y and thus defines a smooth bijective map of S1×S1 onto the torus in 4 space.
