If two simple closed curves intersect transversally at exactly one crossing point in a torus, are they necessary meridian and the longitude of the torus or they may be combination of these two generators?
I think they must be the two generators of the torus (meridian and the longitude), otherwise the points of intersection must be more than one, right?
On
Here's an totally explicit construction of an example:
Let $$\pi : \Bbb R^2 \to (\Bbb R / \Bbb Z) \times (\Bbb R / \Bbb Z) \cong \Bbb S^1 \times \Bbb S^1, \qquad (x, y) \mapsto (x + \Bbb Z, y + \Bbb Z) \sim (e^{2 \pi i x}, e^{2 \pi i y}),$$ be the usual covering of the torus. Then, the images of the $x$- and $y$-axes under $\pi$ are a parallel $P$ and a meridian $M$.
Now, the image of the line $x = ny$, $n \neq 0$, intersects $P$ only at the image $O$ of the origin, but it is not homotopic to $M$ (for either parameterization). It is homotopic, however, to a loop that starts at $O$, traces around $M$ once, and then traces around $P$ $n$ times in the 'positive' direction (or $-n$ times in the 'negative' direction). For $n = 1$, this gives the construction in Anubhav's excellent answer.
A geometric explanation would be..If you cut the torus along one curve, you'll end up with an annulus and since twose two loops intersect only once, if you cut along the other loop you'll get a square...so now you can see the picture that this is (upto a homeomorphism) meridian and longitudinal... so they are generator of the torus as well.
But they are not necessarily homotopic to meridian and longitudinal...for example, starting with meridian and longitudian loop, and apply a Dehn twist along the meridian loop...then you'll endup with one meridian and one other curve which is not homotopic to the longitudinal anymore.