In general we speak of maps from X to Y.
The image of such a map constitutes an embedding of X in Y if distinct points map to distinct points.
If f0 and f1 both map X into Y then they are said to be homotopic if there is a map F of
[0, 1] × X into Y such that F(0, x) = f0(x) and F(1, x) = f1(x) for all x in X. The function F(a, x), for some particular a, need not be an embedding, for we do not require that F(a, x1) be different from F(a, x2). Thus for some a, F(a, x) can map to a single point. Homotopy is an equivalence relation.
If we choose a point y in Y thru which images of two loops pass and in a neighborhood of y change into then we have a new loop which is perhaps in an equivalence class different from either of the input loops. This operation defines a group on the homotopic equivalence classes. If Y is arcwise connected then you can always find representatives of any equivalence class thru any point. This notion pays attention to the orientation of the loop. Two rubber bands circling a broomstick in opposite directions cancel each other out when thus multiplied, and the result is homotopic to a point. They are inverses in the homotopy group.
Wrong!!! The preceding paragraph fell into a trap.
It turns out that it makes a difference homotopically which point you pick to make the splice.
Different choices produce results that are not homotopic to each other!
The easiest counter example is on the punctured torus.
We expand the puncture until it becomes the rectangular border in the pictures below.
In the picture the two dark disks are where the handle recedes behind the plane out of view.
If A is a loop that encircles one end of the handle then it is possible to compute A−1A to be either homotopic to a point or to a loop about the puncture.
These are not homotopic.
Some think that if they know the homotopy group of Y then they know Y. Here is some evidence. Here is a very good book on Algebraic Topology.