The first thing we're going to do is allow the base point c to move during the homotopy. In other words, the loop can venture far afield, as long as it returns to c.
Assume f and g are homotopic loops that happen to be based at c. Follow the base point as the homotopy proceeds. This is a continuous function that starts and ends in c; in other words, a loop. Call this loop h(I), since it is a path induced by the homotopy. We can now build a homotopy that fixes c, and connects f with hg/h. As f moves with time, prepend a path that travels partway along h, from c out to the base point of f. Once f has been traversed, append the return path back to c. At all times, c remains fixed. At the end of the homotopy, the image of our circle runs around h, then g, then back along h. Within the fundamental group, f and g are conjugate.
Conversely, let hf/h represent the conjugate of f by h in π1(S). Apply a general homotopy to hf/h, which retracts the path h on either side, leaving f. This homotopy moves the base point, and returns it to c, but that's ok.
In summary, two loops based at c are homotopic in S, iff they represent conjugates in the fundamental group.
If S is path connected, we can break away from the base point altogether. If g is based at some other point b, let a homotopy gradually extend a path from b back to c. At the far end of the homotopy we have a loop that goes from c to b, around g, and back to c. thus every loop is homotopic to a loop based at c. The loop classes at c are the loop classes for all of S. Let's look again at the loop classes at c.
If π1(S) is abelian, as it often is, then conjugation by h doesn't mean a thing. thus f and g are homotopic in S iff f = g in the fundamental group. This is the case with the circle. It's fundamental group is Z, which is abelian, and its homotopy classes are also Z.
If π1(S) is nonabelian, then hf/h could indeed be different from f. The set of group elements that are conjugates of f is called the conjugacy class of f, and the homotopy classes of S are the conjugacy classes of π1(S). Note that this is a set, not a group, as the group operator does not extend to conjugacy classes.
If S is such a space, then any loop shrinks to a point, and then follows a path back to a base point c. There is but one homotopy class, represented by the constant function c.
As we saw above, the homotopy classes of S are the conjugacy classes of π1(S). If f represents a class in the fundamental group, then ff/f is trivial, which means f is trivial, and π1(S) = 1.
conversely, a trivial fundamental group has but one conjugacy class, whence every loop shrinks to a point, and S is simply connected.
In summary, S, a path connected space, is simply connected iff π1(S) is trivial.
Assume S is not path connected. We say S is simply connected if each path component is simply connected. A loop isn't going to cross between two path components, so each loop exists within one path component, and can shrink to a point.
A space is locally simply connected if every point x in an open set U is contained in an open set Q inside U that is simply connected. The plane without the origin is locally simply connected, but not simply connected.
A space can be simply connected without being locally simply connected. Start with the descending earring, a shape in two dimensions, and let each circle act as the base of a cone that extends out into the third dimension. These cones all meet at a common apex. Any path, including a loop, can slide up to the common apex, hence this space is simply connected. However, an open set containing the origin, and excluding the apex, proves the space is not locally simply connected.
The space S is semilocally simply connected if, for every point x, there is an open neighborhood Q containing x, such that every loop in Q is homotopic to a point, using any homotopy in S. In other words, the loop can leave Q, and then return, as it shrinks to a point. If there is such a Q, we can intersect it with any neighborhood U about x, and the resulting open set still satisfies our criteria. (Forcing Q inside U is the traditional definition of "locally".)
Let f(S) into T, and g(T) into S, implement the equivalence. Let a be the base point of S, and let c = f(a), and let b = g(c). Thus the composite function fg moves the base point from a to b within S. A followup homotopy h slides b back to a, as it morphs g(f(S)) into the identity map on S. Let l be the path from a to b, as defined by this homotopy.
Let u be a loop based at a, and map it forward and back through f and g, giving a loop v based at b. An isomorphism, like the one described in the previous section, turns v into a corresponding loop based at a. Use the path l when building this isomorphism. Thuse v → lv/l. Finally let h slide v along l, shrinking l as we go, producing u. Thus u and lv/l belong to the same homotopy class.
Three functions combine to produce the identity map on the fundamental group. These are f, g, and the base point isomorphism from b back to a along l. Each of these functions is injective, i.e. a monomorphism, else some loop u would not map back to its own class. Furthermore, the third function is an isomorphism. Let v be a loop based at b, and run it through all three functions. This gives a loop w at b that leads to u, and since the base change is an isomorphism, w and v are in the same homotopy class. Therefore g is surjective. By symmetry, f is also surjective. Therefore π1(S) maps uniquely onto π1(T), and the fundamental groups are isomorphic.