Question: Let $M$ and $N$ be $3$-manifolds obtained by zero-surgery on (left-handed) trefoil and figure-eight knot respectively.

What is the easy way to prove that $M$ and $N$ are not homeomorphic?

Note: When they are knot homology spheres (they are both homology $S^1 \times S^2$'s), I cannot use the classical invariants.

Answer: 
If you're happy bringing in heavy machinery then you could compute some sort of Floer homology, like the 'hat' version of Heegaard Floer homology: this has rank 2 for $S^3\_0(3\_1)$ and rank 4 for $S^3\_0(4\_1)$, so they're different.

On the other hand, in cases like this where you have a very specific pair of 3-manifolds in mind, it's often enough to distinguish their fundamental groups by counting covers of some finite order. Here you could work this out by hand starting from Wirtinger presentations of the respective knot groups, or you could just ask SnapPy to do it:

```
In[1]: len(Manifold('3_1(0,1)').covers(9))
Out[1]: 4

In[2]: len(Manifold('4_1(0,1)').covers(9))
Out[2]: 2

```

This counts 9-fold covers of each 0-surgery, and since the numbers are different, they must have different fundamental groups.
This is slightly simpler than the arguments above. The double covers of (0,1) surgeries for these two manifolds have different homologies.

The double cover of the (0,1) surgery on the figure 8 knot complement has homology Z/5Z + Z. The double cover of the (0,1) surgery on the trefoil knot complement has homology Z/3Z + Z.

It is unclear how well distinguishing manifolds by the homologies of their covers works and it is known to fail for certain Sol torus bundles. However, when it succeeds, it provides a concrete invariant for distinguishing manifolds.

Also, the double cover of any two bridge knot complement S^3\TBL(p,q) is the complement of a null-homologous knot complement in L(p,q). Hence it has homology Z/pZ + Z as does the double cover of (0,1) surgery (I am implicitly using that the (0,1) curve lifts in cyclic covers). In this case, the trefoil is the (3,2) two bridge knot and the figure eight knot is the (5,2) two bridge knot. So this is an effective technique for distinguishing many pairs of (0,1) surgeries on two-bridge knot complements.