We can regard $\pi_1(X,x_0)$ as the set of basepoint-preserving homotopy classes of maps $(S^1,s_0)\rightarrow(X,x_0$). Let $[S^1,X]$ be the set of homotopy classes of maps $S^1\rightarrow X$, with no conditions on basepoints. Thus there is a natural map $\Phi:\pi_1 (X,x_0)\rightarrow[S^1,X]$ obtained by ignoring basepoints. Show that $\Phi$ is onto if $X$ is path-connected, and that $\Phi([f])=\Phi([g])$ iff $[f]$ and $[g]$ are conjugate in $\pi_1(X,x_0)$. Hence $\Phi$ induces a one-to-one correspondence between $[S^1,X]$ and the set of conjugacy classes in $\pi_1(X)$, when $X$ is path-connected.
We think of $\pi(X, X_0)$ as homotopy classes of basepoint preserving maps $(S^1, s_0) \rightarrow (X, x_0)$. Recall that a map is basepoint preserving iff $f(s_0) = x_0$.
Define $[S^1, X]$ to be homotopy classes of maps $S^1 \rightarrow X$ with no condition on basepoints.
The exercise asks us to show that the map $\Phi: \pi_1(X, x_0) \rightarrow [S^1, X]$ is an onto map if $X$ is path-connected.
I don't think is true. Consider a wedge of circles. This is path-connected. Now consider all loops based at $x_0$, a point on the left circle that is not on the right circle. In the figure, this is in blue. These loops are in $\pi(X, x_0)$. Now, on the other hand, consider a loop on the circle on the right hand side. In the figure, this loop is in pink. It seems impossible that any blue loop can be homotped into a pink loop.
Indeed, by reading ahead, we know that the fundamental group of the wedge of circles is $W \equiv \mathbb Z * \mathbb Z \simeq \langle a, b\rangle$. All loops on the left circle will be of the form $a^n$, while a loop on the right circle will be of the form $b^m$. The map $\Phi$ that produces all the $a^n$ cannot produce a $b^m$, and thus the map $\Phi$ cannot be surjective.
What am I missing?
Best Answer
Loops based at $x_0$ can be (freely) homotopic to the pink loop. As an example consider the loop starting at $x_0$ which travels along the upper half of the left circle until it reaches the intersection of both circles, then travels once along the pink circle and travels back along the upper half of the left circle to $x_0$.