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Let X be a topological space, and let $x_0$ be a point of X. Show that if $\lambda_1$ and $\lambda_2$ in $\pi(X; x_0)$ have the same image under $\pi(X; x_0) \rightarrow [S^1;X]$, then $\lambda_1$ and $\lambda_2$ are conjugate. (Conjugate in the group sense).

I'm confused as to what the images look like in $[S^1; X]$

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Hint: $\pi_1$ is also defined in terms of loops in $X$ — so what's the difference b/w $\pi_1(X)$ and $[S^1,X]$?.. –  Grigory M Nov 19 '11 at 18:34

1 Answer 1

I think this question is really of the following form: if $f_1,f_2: S^1 \to X$ represent elements $\alpha, \beta$ in $G=\pi_1(X,x_0)$, and $f_1,f_2$ are freely homotopic, then $\alpha, \beta$ are conjugate in the group $G$. Here freely homotopic means not respecting the base point.

So if $a$ is the base point of $S^1$, then $f_1(a)=f_2(a)=x_0$, but the homotopy determines an element $\gamma$ of $G$ represented by the path, say $k$, of the base point during the homotopy. Then $\gamma$ conjugates $\alpha$ to $\beta$ in $G$.

There are several ways to see this. One is to represent the homotopy as a square in which the bottom and top edges are given by $f_1,f_2$ and the vertical edges are given by $k$, since the square identifies to a cylinder to give a homotopy of maps of $S^1$. Because the square is convex, it determines a homotopy which gives the equation $\alpha \gamma= \gamma \beta$.

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