Given a map f from S cross I into Y, build a map g from S into Z as follows. For any x in S, let g(x) be the path defined by f restricted to x. In other words, x establishes a closed interval in the product space S cross I, and f carries this interval to a path in Y, which is a point in Z. Conversely, a map g(S) into Z pulls back to a map f(S:I) into Y, and the two transformations are inverse to eachother.
Picture this when S is a circle. Cross S with I and get the wall of a cylinder. Map this into Y via a function f. The cylinder can be "unraveled" into vertical line segments, one for each point of S. These segments become paths in Y, which are points of Z. And if S maps into points of Z, it defines paths in Y, giving a map from the cylinder into Y.
We will prove f is continuous iff g is continuous. That's pretty intuitive, but the proof is rather technical.
Start with f continuous and select an open set O in Z. This is defined by finitely many pairs K and U, where paths map the compact set K in I into the open set U in Y. Let p be such a path, a point in O, and let q be a point of S such that g(q) = p. If you think of the earlier example, q is a point on the circle, and g maps q into a path that carries K into U. Equivalently, slide up the wall of the cylinder starting at q. As you pass through the regions of K, f maps that portion of the cylinder into U.
Since f is continuous, let W be the open preimage of U. This is a smear on the wall of the cylinder, containing K. Since K and q are both compact, K cross q is also compact. Thus K (on the wall) is covered by finitely many base open sets inside W. Each of these is an open set in S cross an open interval in I. for convenience, let T be the intersection of the open sets in S. Now T is still open, and can be crossed with the open intervals in I to cover K.
Consider g(T). Each point in the image is a path in Y that maps K into U. In other words, g(T) lies in O - and T includes q. Intersect this result over finitely many K and U, and the preimage of O is open in S. Therefore g is continuous.
Conversely, assume g is continuous. Let U be an open set in Y. Let p be a point in U with preimage q cross x somewhere on the wall of the cylinder. The path defined by q is a point in Z. It is also a path in Y that maps x into U. Since paths are continuous, this particular path maps some closed interval K, containing x, into U.
Since K is compact, an open set in Z consists of functions that map K into U. This has an open preimage under g. Therefore some open set T in S exists, such that f carries T cross K into U. Remove the endpoints of K, and f maps the open set T cross K into U. Therefore f is continuous.
In summary, the unraveling transformation carries continuous functions uniquely onto continuous functions.
This seems like it's going to be a nightmare, but maybe not. A homotopy is a continuous function from (S*I)*J into Y, where J is another unit interval, used to build the homotopy. Yet the product of finitely many topological spaces is the same, up to isomorphism, regardless of order. So write the domain as (S*J)*I. Unravel this continuous function to get another one, from S*J into Z. The boundary conditions are satisfied, and the result is a homotopy.
Reverse the above to carry a homotopy in Z back to a homotopy in Y.
With S cross I as domain, consider only those functions that map S cross 0, S cross 1, and d cross I onto e. Continuing the cylinder example, the bottom and top circles map to e, and the vertical line at d also maps to e. If the bottom circle, the top circle, and the vertical line are compressed down to a single point, no information is lost. All these things mapped to e anyways. Therefore, let H be the suspension of S cross I, and let f map H into Y, with f(d) = e. The counterpart g takes d to the trivial loop e in Z. Review quotient spaces and note that f is continuous on H iff it is continuous on S cross I, iff g is continuous on S.
Finally verify the homotopy classes are the same. A homotopy on H, mapping d cross J onto e, is the quotient space of a homotopy on S cross I, mapping the appropriate boundary sets to e. Unravel the homotopy on S cross I into Y, giving a homotopy on S into Z. Within this unraveled homotopy, the image of d is always the trivial loop e. In other words, we have a base point homotopy. And this process can be reversed, giving a homotopy on S cross I, which becomes a homotopy on H courtesy of a quotient map. In summary, images of H correspond to images of S, and homotopy classses of H correspond to homotopy classes of S, such that d always maps to e.
The set of homotopy classes of H into Y (mapping d to e) corresponds to the homotopy classes of S into Z (mapping d to the constant loop e). Put another way, πn(Y) = πn-1(ω(Y)).
By induction, pull πn(Y) down to π1 of the n-1 order loop space of Y. Since the fundamental group is, well, a group, we can impose the same group structure on πn(Y). Furthermore, for n > 1, πn is abelian. This follows from the previous theorem, which states the fundamental group of a loop space is abelian.