Loops, Base Points, and Simple Connectivity
The fundamental group records loops based at a chosen point, up to homotopies that keep that point fixed. Simple connectivity expresses a global absence of nontrivial loops, together with the ability to join any two points by a path. The path-connectedness condition matters: a space can have trivial fundamental groups at all of its points and still consist of more than one path component.
Let \(I=[0,1]\) with its usual topology, and let \(X\) be a topological space. A path from \(x_0\) to \(x_1\) is a continuous map \(\gamma:I\to X\) with \(\gamma(0)=x_0\) and \(\gamma(1)=x_1\). A loop based at \(x_0\) is a continuous map \(\alpha:I\to X\) with \(\alpha(0)=\alpha(1)=x_0\). Write \(c_{x_0}\) for the constant loop at \(x_0\).
The fundamental group \(\pi_1(X,x_0)\) is the group of based loops at \(x_0\) modulo based homotopy; its identity element is the class \([c_{x_0}]\) of the constant loop. Thus the relationship between null-homotopy and the identity class is built directly into the definitions, not inferred from the shape of a loop.
Proof. By definition, two based loops represent the same element of \(\pi_1(X,x_0)\) exactly when a homotopy between them keeps both endpoints fixed at \(x_0\). Therefore \([\alpha]=[c_{x_0}]\) holds exactly when such a homotopy joins \(\alpha\) to \(c_{x_0}\), which is precisely a based null-homotopy. \(\square\)
The Fundamental-Group Characterization
- \(X\) is simply connected.
- Every based loop at every point of \(X\) is null-homotopic as a based loop.
- \(\pi_1(X,x)\) is the trivial group for every \(x\in X\).
- \(\pi_1(X,x_0)\) is the trivial group for at least one \(x_0\in X\).
Proof. Statements (1) and (2) are equivalent by the definition of simple connectivity, since path-connectedness is assumed. By the proposition, at any chosen \(x\in X\), every based loop at \(x\) is null-homotopic if and only if every element of \(\pi_1(X,x)\) is its identity. Applying this at every \(x\) proves the equivalence of (2) and (3).
Statement (3) implies (4) because \(X\) is nonempty: a path-connected space has at least one point, so any choice of \(x_0\in X\) has trivial fundamental group.
For (4) to imply (3), use path-connectedness. The base-point-change theorem says that if \(\gamma\) is a path from \(x_0\) to \(x\), then \(\pi_1(X,x_0)\) and \(\pi_1(X,x)\) are isomorphic. In particular, the isomorphism sends the identity group to the identity group: an isomorphism from a group with one element has an image with one element and is onto. Since \(\pi_1(X,x_0)\) is trivial, \(\pi_1(X,x)\) is trivial for every \(x\in X\). This proves (3), and hence all four statements are equivalent. \(\square\)
The theorem gives two useful ways to check simple connectivity. One may work directly with a based homotopy for each loop, or work with the fundamental group at one base point and use path-connectedness to transfer the conclusion. The second method is often shorter when a fundamental group has already been computed.
Worked Example: Every Loop in the Interval Is Null-Homotopic
Consider \(I=[0,1]\) with the subspace topology inherited from \(\mathbb{R}\), and choose any base point \(x_0\in I\). Let \(\alpha:I\to I\) be any loop based at \(x_0\). Define
For each \(s,t\in I\), both \(\alpha(t)\) and \(x_0\) lie in \([0,1]\). Since \(0\leq s\leq1\), the coefficients \(1-s\) and \(s\) are nonnegative and sum to \(1\). Thus \(H(s,t)\) lies between \(\alpha(t)\) and \(x_0\), so \(H(s,t)\in[0,1]\). The formula uses continuous real-valued operations and the continuous map \(\alpha\), so \(H\) is continuous.
At \(s=0\), \(H(0,t)=\alpha(t)\), and at \(s=1\), \(H(1,t)=x_0=c_{x_0}(t)\). At either endpoint of the loop parameter, \(\alpha(0)=\alpha(1)=x_0\), so \(H(s,0)=(1-s)x_0+sx_0=x_0\) and \(H(s,1)=(1-s)x_0+sx_0=x_0\). Therefore \(H\) is a based null-homotopy. This works for every loop and every base point; the interval is path-connected, so it is simply connected.
Check with a specific loop: take \(x_0=0\) and \(\alpha(t)=4t(1-t)\). Its endpoint values are \(\alpha(0)=4\cdot0\cdot1=0\) and \(\alpha(1)=4\cdot1\cdot0=0\), and \(0\leq4t(1-t)\leq1\) for \(t\in[0,1]\). At \(t=\tfrac12\), \(\alpha(\tfrac12)=4\cdot\tfrac12\cdot\tfrac12=1\). The homotopy is \(H(s,t)=(1-s)4t(1-t)\); in particular \(H(\tfrac12,\tfrac12)=(1-\tfrac12)\cdot1=\tfrac12\). For all \(s\), \(H(s,0)=H(s,1)=0\), as required.
Worked Example: A Based Contraction in the Plane
Give \(\mathbb{R}^2\) its usual Euclidean topology. Consider the loop based at \((0,0)\) given by \(\alpha(t)=(\sin(2\pi t),1-\cos(2\pi t))\). At \(t=0\) its value is \((\sin 0,1-\cos 0)=(0,0)\), and at \(t=1\) its value is \((\sin 2\pi,1-\cos 2\pi)=(0,0)\), so it is indeed a based loop.
Set \(H(s,t)=((1-s)\sin(2\pi t),(1-s)(1-\cos(2\pi t)))\). Each coordinate is continuous in \(s,t\), so \(H:I\times I\to\mathbb{R}^2\) is continuous. At \(s=0\) it equals \(\alpha(t)\), and at \(s=1\) it equals \((0,0)\) for every \(t\). At \(t=0\) and \(t=1\), both coordinates are zero for every \(s\); hence the base point remains fixed.
For a numerical check, at \(t=\tfrac14\) we have \(\sin(2\pi t)=\sin(\pi/2)=1\) and \(1-\cos(2\pi t)=1-\cos(\pi/2)=1\). Thus \(\alpha(\tfrac14)=(1,1)\), and \(H(\tfrac12,\tfrac14)=(\tfrac12,\tfrac12)\). At the end of the homotopy, \(H(1,\tfrac14)=(0,0)\), while \(H(s,0)=H(s,1)=(0,0)\) for every \(s\). This verifies the contraction conditions for this particular loop.
Why Path-Connectedness Cannot Be Dropped
Worked Example: Trivial Fundamental Groups Without Simple Connectivity
Let \(X=\{a,b\}\) have the discrete topology, in which every subset is open. A continuous path \(\gamma:I\to X\) cannot join \(a\) to \(b\). Indeed, if such a path existed, \(\gamma^{-1}(\{a\})\) and \(\gamma^{-1}(\{b\})\) would be disjoint, nonempty open subsets of \(I\) whose union is \(I\); each would also be closed because its complement is open. This would separate \(I\), contradicting the connectedness of the interval. Thus \(X\) is not path-connected.
Every loop in \(X\) is constant. To verify this, a continuous image of the connected space \(I\) is connected, whereas any subset of a discrete space with at least two points is disconnected: a singleton and its complement give a separation. Hence a loop has a one-point image. Consequently the only based loop at \(a\) is \(c_a\), and the only based loop at \(b\) is \(c_b\). Therefore \(\pi_1(X,a)\) and \(\pi_1(X,b)\) each have exactly one element. The space is nevertheless not simply connected, because it is not path-connected.
Common pitfall. A tempting but incorrect conclusion is: “\(\pi_1(X,a)\) is trivial, so \(X\) is simply connected.” For the discrete space just computed, \(\pi_1(X,a)=\{[c_a]\}\), but there is no path from \(a\) to \(b\). The missing hypothesis is path-connectedness. The correct conclusion from triviality at one base point is simple connectivity only when the space is also path-connected; under that hypothesis, the base-point-change theorem transfers triviality to every point.
This distinction is useful whenever a fundamental group is used to describe a whole space. A fundamental group is attached to a chosen base point, and path-connectedness is what lets information at that point reach the other base points. Without it, triviality can describe each component’s loops while saying nothing about whether different components can be joined.
Check Your Understanding
Use the based-homotopy definition and the fundamental-group characterization in your answers.
- State all four equivalent conditions in the theorem, and identify where path-connectedness is used.
- Explain why a based null-homotopy of \(\alpha\) implies \([\alpha]=[c_{x_0}]\).
- For the interval homotopy \(H(s,t)=(1-s)\alpha(t)+sx_0\), verify explicitly what happens at \(s=0\), \(s=1\), \(t=0\), and \(t=1\).
- Why does the discrete space \(\{a,b\}\) have trivial fundamental groups but fail to be simply connected?
- If \(X\) is path-connected and \(\pi_1(X,x_0)\) is trivial at one point, what established result lets you conclude that \(\pi_1(X,x)\) is trivial at every point?