The concept of a loop plays an essential role to characterize simple connectedness. Let us begin by introducing an equivalence relation for loops $u, v \in \mathcal L'(X)$ which is coarser than path homotopy. Recall that $u \simeq_p v$ requires that $u, v$ are based at the same point $x_0 \in X$ and $H : u \simeq_p v$ is stationary on $\{0,1\}$.
Given loops $u,v \in \mathcal L'(X)$, a loop homotopy is a homotopy $H : I \times I \to X$ from $u$ to $v$ such that for each $s \in I$ the map $H_s : I \to X, H_s(t) = H(t,s)$, is a loop. In that case we say that $u,v$ are loop homotopic, $u \simeq_l v$. Clearly $\simeq_l$ is an equivalence relation on $\mathcal L'(X)$ and $u \simeq_p v$ implies $u \simeq_l v$.
Let us next give an alternative interpretation of a loop.
Loops can be identified with maps $S^1 \to X$ living on the circle $S^1 = \{ z \in \mathbb C : \lvert z \rvert = 1 \}$. In fact, let $\mathcal L(X)$ denote the set of all these maps (which is denoted as the free loop space of $X$) and let $\exp : I \to S^1, \exp(t) = e^{2\pi i t} = \cos(2\pi t) + i \sin(2\pi t)$. This is a closed surjection, hence a quotient map. For $t, t' \in I$ we have $\exp(t) = \exp(t')$ if and only if $t = t'$ or $\{t, t'\} = \{0 ,1 \}$. Thus it identifies $0,1 \in I$ to a single point. Therefore $\exp^* : \mathcal L(X) \to \mathcal L'(X), \exp^*(f) = f \circ \exp$, is a bijection. To $S^1$ we give $1$ as a basepoint and say that $f : S^1 \to X$ is based at $x_0 \in X$ if $f(1) = x_0$.
Under this bijection loop homotopy in $\mathcal L'(X)$ corresponds to ordinary homotopy of maps in $\mathcal L(X)$. Path homotopy in $\mathcal L'(X)$ corresponds to basepoint-preserving homotopy in $\mathcal L(X)$, where a basepoint-preserving homotopy is required to be stationary on the basepoint $1 \in S^1$. This is true because the map $\exp \times id_I : I \times I \to S^1 \times I$ is a closed surjection and thus a quotient map.
Here are our results.
For a path-connected space $X$ the following are equivalent:
|
Path form |
Circle form |
| 1 |
Each loop $u : I \to X$ is path homotopic to the constant loop based at $x_0 = u(0) = u(1)$. |
Each $f : S^1 \to X$ is basepoint-preserving homotopic to the constant map based at $x_0 = f(1)$. |
| 2 |
Any two loops $u, v : I \to X$ based at the same $x_0 \in X$ are path homotopic. |
Any two $f, g : S^1 \to X$ based at the same $x_0 \in X$ are basepoint-preserving homotopic. |
| 3 |
Any two paths $u, v : I \to X$ with the same endpoints are path homotopic. |
|
| 4 |
Each loop $u : I \to X$ is loop homotopic to a constant loop in $X$. |
Each $f : S^1 \to X$ is homotopic to a constant map $S^1 \to X$. |
| 5 |
|
Each $f : S^1 \to X$ has a continuous extension $F : D^2 \to X$. |
For an arbitrary space $X$ simple connectedness is equivalent to each of the following properties:
|
Path form |
Circle form |
| 6 |
Each loop $u : I \to X$ and each constant loop in $X$ are loop homotopic. |
Each $f : S^1 \to X$ and each constant map $S^1 \to X$ are homotopic. |
| 7 |
Any two loops $u, v : I \to X$ are loop homotopic. |
Any two $f, g : S^1 \to X$ are homotopic. |
It is clear that path and circle form in the same line are equivalent.
Of course (6) and (7) are valid characterizations of simple connectedness also in the special case that $X$ is path-connected. The point is that they imply path-connectedness. This will be shown below.
(2) $\Rightarrow$ (1) : Obvious, (1) is a special case of (2).
(1) $\Rightarrow$ (2) : This follows from the fact that $\simeq_p$ is an equivalence relation.
(1) $\Rightarrow$ (4) : Obvious, (4) is a special case of (2).
(4) $\Rightarrow$ (5) : Let $H : S^1 \times I \to X$ be a homotopy such that $H_0 = f$ and $H_1 = c$, where $c$ is a constant map (with value $x_0 \in X$). The map $p : S^1 \times I \to D^2, p(z,t) = (1-t)z$, is a closed surjection, hence a quotient map. It maps $S^1 \times \{1\}$ to $0 \in D^2$. Therefore $H$ induces a unique function $F : D^2 \to X$ such that $F \circ p = H$. Note that $F(z) = H(z/\lvert z \rvert,1 - \lvert z \rvert)$ for $z \ne 0$ and $F(0) = x_0$. By the universal property of quotient maps, $F$ is continuous. For $z \in S^1$ we have $F(z) = H(z,0) = f(z)$, thus $F$ is the desired continuous extension of $f$.
(5) $\Rightarrow$ (1) : Let $F : D^2 \to X$ be a continuous extension of $f$. Define $H : S^1 \times I \to X, H(z,t) = F((1-t)z +t)$. This is a well-defined continuous map (note that $\lvert (1-t)z +t \rvert \le (1-t)\lvert z \rvert + t = 1 - t + t = 1$). We have $H(z,0) = F(z) = f(z), H(z,1) = F(1) = f(1) = x_0$ and $H(1,t) = F(1) = f(1) = x_0$.
(3) $\Rightarrow$ (2) : Obvious, (2) is a special case of (3).
(5) $\Rightarrow$ (3) : Let $x_0 = u(0) = v(0)$ and $x_1 = u(1) = v(1)$. Let $S = I \times \{0,1\} \cup \{0,1\} \times I$ denote the boundary of the square $D = I \times I$. Points of $S$ will be written in the form $(t,s)$. Define
$$\phi : S \to X, \begin{cases} u(t) & s = 0 \\ v(t) & s = 1 \\ x_0 & t = 0 \\ x_1 & t = 1 \end{cases}$$
Our aim is to find a continuous extension $H : D \to X$ of $\phi$; this is the desired path homotopy from $u$ to $v$. To do so, it suffices to find a homeomorphism $h : D \to D^2$ such that $h(S) = S^1$; then (6) can be applied to find a continuous extension $F : D^2 \to X$ of $f = \phi \circ h^{-1}$. Then clearly $H = F \circ h$ is the desired extension of $\phi$. The existence of such $h$ is well-known. We shall nevertheless give a proof in the end of this answer.
Remark. Proving (5) $\Rightarrow$ (1) is basically superfluous because it is covered by (5) $\Rightarrow$ (3) $\Rightarrow$ (2) $\Rightarrow$ (1). However, the short and transparent proof of (5) $\Rightarrow$ (1) is interesting in its own right.
Let us now come to the characterizations (6) and (7) of simple connectedness of arbitrary spaces $X$.
$X$ simply connected $\Rightarrow$ (6) : If $X$ is simply connected, then it is path connected. Thus (6) follows from (4) because in a path connected space any two constant maps $S^1 \to X$ are homotopic.
(6) $\Rightarrow$ (7) : Obvious because loop homotopy is an equivalence relation.
(7) $\Rightarrow$ (6) : Obvious, (6) is a special case of (7).
(6) $\Rightarrow$ $X$ simply connected : Let $a, b \in X$. The constant loops based at $a$ and $b$ are loop homotopic via a homotopy $H: I \times I \to X$. Then $u(s) = H(0,s)$ is a path from $a$ to $b$. This means that $X$ is path connected. Thus it suffices to show that (6) implies (4). This is obvious, pick any constant map $S^1 \to X$.
For the sake of completeness we finally construct a homeomorphism $h : D \to D^2$ such that $h(S) = S^1$.
Let $J = [-1,1]$ and $\alpha : I \to J, \alpha(t) = 2t - 1$. This is a homeomorphism; hence $\beta = \alpha \times \alpha : D \to \bar D = J \times J$ is a homeomorphism such that $\beta(S) = \bar S = J \times \{-1,1\} \cup \{-1,1\} \times J = $ boundary of $\bar D$. Note that $\bar D = \{ (t,s) \in \mathbb R^2 : \lVert (t,s) \rVert_\infty = \max(\lvert t \rvert, \lvert s \rvert) \le 1 \}$ and $\bar S = \{ (t,s) \in \mathbb R^2 : \lVert (t,s) \rVert_\infty = 1 \}$. Also note that $\lvert t \rvert, \lvert s \rvert \le \sqrt{t^2 + s^2}$, thus $\lVert (t,s) \rVert_\infty \le \lVert (t,s) \rVert = $ Euclidean norm of $(t,s)$. Now define
$$\bar h : \bar D \to D^2, \bar h(t,s) = \begin{cases} \dfrac{\lVert (t,s) \rVert_\infty}{\lVert (t,s) \rVert_\phantom{\infty}}(t,s) & (t,s) \ne (0,0) \\ (0,0) & (t,s) \ne (0,0) \end{cases}$$
This map is well-defined because $\left\lVert \dfrac{\lVert(t,s) \rVert_\infty}{\lVert (t,s) \rVert_\phantom{\infty}}(t,s) \right\rVert = \lVert(t,s) \rVert_\infty \le 1$. It is clearly continuous in all $(t,s) \ne (0,0)$. Continuity in $(0,0)$ follows from $\lVert \bar h(t,s) - \bar h(0,0) \rVert = \left\lVert \dfrac{\lVert(t,s) \rVert_\infty}{\lVert (t,s) \rVert_\phantom{\infty}}(t,s) \right\rVert = \lVert(t,s) \rVert_\infty \le \lVert(t,s) \rVert$. Similarly
$$h^* : D^2 \to \bar D, h^*(t,s) = \begin{cases} \dfrac{\lVert (t,s) \rVert_\phantom{\infty}}{\lVert (t,s) \rVert_\infty}(t,s) & (t,s) \ne (0,0) \\ (0,0) & (t,s) \ne (0,0) \end{cases}$$
is well-defined and it has the property $h^* \circ \bar h = id , \bar h \circ h^* = id$. Therefore $h$ is a bijection. But continuous bijections between compact Hausdorff spaces are homeomorphisms. Finally note that $\bar h(\bar S) = S^1$. Then $h = \bar h \circ \beta$ is the desired homeomorphism.
