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I am reading a proof about the fact that the fundamental group $\pi_1({G,1})$ is abelian if $G$ is a topological group.

The proof starts by taking two loops $f,g$ at $1$ to define the function $H:I^2\to G, H(s,t)=f(s)g(t).$

The autor say that "$H$ sends the boundary of the square (starting at $(0, 0)$ and going counterclockwise) to the concatenation $f·g·\overline{f}·\overline{g}$.

Ok we have $f(0)=f(1)=g(0)=g(1)=1$ so $H(0,0)=H(1,0)=H(1,1)=H(0,1)$ so the assertion follow.

But I don't understand the next step: "Since $H$ is defined continuously on the whole square, it follows that $[f][g][f]^{−1}g]^{−1}$ is the identify in $\pi_1(G,1)$."

I don't understand the part when he said it follows that.

Alex
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1 Answers1

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In general, if $H\colon I^2\to X$ is any continuous map, then the loop defined by $H$'s values on the boundary of the square is homotopic to the constant loop. To write down such a homotopy, let $K(t,x,y) = H(tx,ty)$ for $0\le t\le 1$; then for each such $t$, the boundary of the square-of-side-length-$t$ is an intermediate loop between the original boundary loop and the constant loop $H(0,0)$.

Greg Martin
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