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A topological space $X$ is simply connected if it is pathwise connected and each closed path $u : I = [0,1] \to X$ is path homotopic to the constant path at $x_0 = u(0) = u(1)$.

Recall that

  1. A closed path $u$ is one with $u(0) = u(1)$. A closed path will also be called a loop and we write $\mathcal L'(X)$ for the set of all loops in $X$. If we want to be more precise, we say $u$ is a loop based at $x_0 \in X$ provided $x_0 = u(0) = u(1)$. By $\mathcal L'(X,x_0)$ we denote the set of all loops in $X$ which are based at $x_0$. Also note that each constant path is a loop.

  2. Two paths $u, v : I \to X$ with the same initial point $x_0 = u(0) = v(0)$ and the same terminal point $x_1 = u(1) = v(1)$ are path homotopic, $u \simeq_p v$, if there exists a homotopy $H : I \times I \to X$ such that $H(t,0) = u(t)$ and $H(t,1) = v(t)$ for all $t$ and $H(0,s) = x_0, H(1,s) = x_1$ for all $s$ (such a homotopy $H$ is stationary on the boundary $\partial I = \{0,1\}$; it is called a path homotopy). It is well-known that $\simeq_p$ is an equivalence relation on the set $\mathcal P(X)$ of all paths in $X$. Clearly $\simeq_p$ restricts to an equivalence relation on $\mathcal L'(X)$ and to an equivalence relation on each $\mathcal L'(X,x_0)$.

  3. The set of equivalence classes $\mathcal L'(X,x_0)/\simeq_p$ is nothing else than the fundamental group $\pi_1(X,x_0)$ with base point $x_0$.

Question: Find a list, as complete as possible, of properties characterizing topological spaces which are simply connected.

Each of these properties can then be taken as an alternative definition of simply connected.

Note that many questions in this forum deal with aspects of simple connectedness. The answer to the present question will hopefully cover most of them.

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    $\begingroup$ A superfluous by quite neat characterisation for a space $X$ to be simply connected is that it admit no essential map $K\rightarrow X$ from a CW complex $K$ of dimension $\leq 1$. Clearly this includes the cases $K=S^0,S^1$, and so neatly avoids having to explicitly mention paths at all. Of course you can formulate this instead with simplicial complexes, etc... $\endgroup$
    – Tyrone
    Commented Mar 1, 2021 at 15:44
  • $\begingroup$ See also math.stackexchange.com/q/3117361. $\endgroup$
    – Paul Frost
    Commented Jun 29, 2021 at 12:44

1 Answer 1

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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 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.
6 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
7 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.
8 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.

(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) : Obvious, in a path connected space any two constant maps $S^1 \to X$ are homotopic.

(5) $\Rightarrow$ (4) : Obvious, pick any constant map $S^1 \to X$.

(4) $\Rightarrow$ (6) : 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 $H'(0) = x_0$. 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 $f$.

(6) $\Rightarrow$ (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).

(6) $\Rightarrow$ (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.

Let us now come to the characterizations (7) and (8) of simple connectedness.

If $X$ is simply connected, then it is path connected and (5) shows that (7) is satisfied. Conversely, let (7) be satisfied and 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 and satisfies (5), hence it also satisfies (1) and is therefore simply connected.

(7) $\Rightarrow$ (8) : Obvious because loop homotopy is an equivalence relation.

(8) $\Rightarrow$ (7) : Obvious, (7) is a special case of (8).

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.

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  • $\begingroup$ This is also valid for maps $S^n \to X$? $\endgroup$ Commented May 20, 2023 at 14:56

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