Does simply connectedness require fixing the endpoints when homotopying to a point?

Question

A space $X$ is simply connected if it is path connected and every loop is homotopic to a constant loop.

A space $X$ is simply connected if it is path connected and has trivial fundamental group.

Thinking about these definitions, the second one seems a priori more strong, since going to the definition of fundamental group, it's essentially saying

A space $X$ is simply connected if it is path connected and every loop is homotopic relative $\{0,1\}$ to a constant loop.

I'd like to know if these definitions are indeed equal because I went bugging for a moment there.

Attempt of proof: let $f:[0,1]\to X$ be a loop (i.e. $f(0)=f(1)$). Suppose $F$ is a homotopy that takes $F_0=f$ to $F_1=c$ a constant loop. Let $$G_s(t)=\begin{cases} F_{3t}(0)\text{ if }0\leq t\leq \frac s 3\\ F_s(\frac {3t-s}{3-2s}) \text{ if }\frac s 3\leq t\leq 1-\frac s 3\\ F_{3(1-t)}(0) \text{ if }1-\frac s 3\leq t\leq 1 \end{cases}.$$

This way $G_0=f$ and $G_1$ goes at $c$, stays there some time and comes back which is of course homotopic to staying at $f(0)$ the whole time. Moreover $G_s(0)$ is always equal to $f(0)$.

Would this work? Otherwise please give a hint!

Notes: to me,

• a loop is either a map from $S^1$ or a map from $[0,1]$ that agrees at $0$ and $1$. Depending on the situation I'll use one or the other definition.
• In the first case, a basepoint fixing homotopy between loops $f$ and $g$ is a homotopy $F$ taking $F_0=f$ to $F_1=g$ such that $F_s(\mathbf{z})$ remains constant across all $s\in [0,1]$, for some $\mathbf{z}\in S^1$.
• In the second case, a homotopy of loops between $f$ and $g$ is a homotopy $F$ such that $F_0=f$; $F_1=g$ and $F_s(0)=F_s(1)$ for all $s\in [0,1]$. $F$ is furthermore called endpoint-fixing if its a homotopy relative to $\{0,1\}$ (so it's a stronger statement obviously).

For instance, in my proof, I preferred to write the loops as closed paths. In other contexts, I'll say they're maps from the circle. If this is not standard, please do tell me.

Answer 1

You did not give the source of the definition

A space $X$ is simply connected if it is path connected and every loop is homotopic to a constant loop.

The crucial part is $$\text{every loop is homotopic to a constant loop} \tag{1}$$ The interpretation of $(1)$ should have been clarified in the unknown source.

Certainly we must not understand it in the sense that for each loop $f : [0,1] \to X$ there exists a homotopy $H : [0,1] \times [0,1] \to X$ such that $H(t, 0) = f(t)$ and $H(t,1) = c$ without any requirements concerning the endpoints. If we allow such free homotopies, then each path connected space would be simply connected which does not make sense. In fact, the homotopy $H(t,s) = f((1-s)t)$ deforms $f$ into the constant loop with value $f(0)$.

The standard interpretation of $(1)$ is that the homotopy $H: F \simeq c$ is a homotopy of paths, i.e. a homotopy rel. $\{0,1\}$.

For a more detailed discussion see Characterizing simply connected spaces.

Answer 2

$\newcommand\pinch{\bigcirc\!-}$You are right to be suspicious, but the notions are indeed equivalent.

Let me recall the definitions. A homotopy between continuous maps $f,g:X\rightarrow Y$ is a continuous map $h:[0,1]\times X\rightarrow Y$ with $h(0,x) = f(x)$ and $h(1,x) = g(x)$ for all $x\in X$. If the spaces $X$ and $Y$ are pointed by some $x_0\in X$ and $y_0\in Y$, and if the maps $f,g$ are basepoint-preserving, such a homotopy is a basepoint preserving homotopy, if $h(s,x_0) = y_0$ for all $s\in [0,1]$.

The fundamental group of $X$ at $x_0$ is the set $\pi_1(X,x_0) = [(\Bbb S^1,1),(X,x_0)]_\ast$ of basepoint preserving loops modulo basepoint preserving homotopy.

Claim 1: If $X$ is path-connected, this doesn't depend on $x_0$.

Indeed, let $x_1\in X$ be another basepoint and $\gamma:x_0\rightarrow x_1$ a path. Consider the quotient $Q := \Bbb S^1/(\Re(z)\geq 0) \cong \Bbb S^1 \cup [1,2]$, where $\Bbb S^1 = \{z\in\Bbb C\mid \vert z\vert = 1\}$. Let $\lambda$ be a loop pointed at $y_0$, ie a basepoint preserving map $\lambda:(\Bbb S^1,1)\rightarrow(X,x_0)$. Then going $\gamma$ then $\lambda$ and then $\gamma^{-1}$ defines a loop pointed at $x_0$, by means of a composite map $\Bbb S^1 \rightarrow Q \xrightarrow{\gamma^{-1}\ast\lambda\ast\gamma} X$. Doing the same with $\gamma^{-1}$ we can turn any loop pointed at $x_0$ into a loop pointed at $y_0$. This defines maps $\pi_1(X,x_1) \rightarrow \pi_1(X,x_0)$ and $\pi_1(X,x_1)\rightarrow\pi_1(X,x_0)$ and one may check that these are actual group-homomorphisms. Writing $\omega_0$ for the loop at $x_0$ given by going $\gamma$ and then $\gamma^{-1}$ we note that $[\omega_0] = [\mathbf{1_{x_0}}]\in\pi_1(X,x_0)$ and thus the composite $\pi_1(X,x_0)\rightarrow\pi_1(X,x_1)\rightarrow\pi_1(X,x_0)$ maps the class $[\lambda]$ of a loop $\lambda$ pointed at $x_0$ to the product of loops $[\omega_0]\ast[\lambda]\ast[\omega_0] = [\lambda]$. This (and the analogous observation for $\pi_1(X,x_1)$) shows that the groups are isomorphic.

Claim 2: The two notions of simply connected coincide.

By Claim 1 it suffices to show that every (unpointed) nullhomotopic loop $\lambda:\Bbb S^1 \rightarrow X$ can be made pointed null-homotopic, meaning $[\lambda]=\mathbf{1_{x_0}}\in\pi_1(X,x_0)$, where $x_0 = \lambda(1)$. So let $h:[0,1]\times\Bbb S^1\rightarrow X$ be a homotopy from $\lambda$ to $\mathbf{1}_{x_1}$ for $x_1 = h(1,1)$. Define $\eta:[0,1]\times[1,2]\rightarrow X,\eta(s,t) = h(s(2-t),1)$. Now consider the homotopy $H:[0,1]\times\Bbb S^1 \rightarrow [0,1]\times Q \cong [0,1]\times(\Bbb S^1 \cup [1,2]) \cong [0,1]\times\Bbb{S}^1 \cup [0,1]\times[1,2] \xrightarrow{h \cup \eta} X$. This map is continuous as composite of continous maps. Furthermore $H(s,1) = \eta(s,2) = h(0,1) = \lambda(1)$ for all $s$, $H(0,z)$ is a reparametrization of $\lambda$ and $H(1,z)$ is a basepoint preserving null-homotopic path at $x_0$. This shows that $[\lambda] = [\mathbf{1}_{x_0}] \in \pi_1(X,x_0)$ as claimed.

Answer 3

The homotopy class of a free loop in $X$ corresponds to a conjugacy class in $\pi_1(X)$, which will be trivial if and only if there are no nontrivial conjugacy classes. Therefore the definitions of simply-connectedness via free loops and based loops are equivalent.