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Show that if $X$ is a topological space and $\pi_1(X, x_0) = 0$, then every map $S^1 \to X$ is null-homotopic.

I think I've boiled this down to the fact that I would need to be sure that if $f : S^1 \to X$ is any map, then $f(S^1)$ is a loop in $X$.

Let $f : S^1 \to X$ be a map such that $x_0 \in f(S^1)$. Now if the image of $f$ is a loop, then $[f] \in \pi_1(X, x_0)$ and since $\pi_1(X,x_0) = 0$ this implies that $f$ is homotopic to $x_0$ i.e. null-homotopic.

I think I need to make these few assumptions here that $f(S^1)$ is a loop and that $x_0 \in f(S^1)$? Are these necessary?

Walker
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    To me, a loop is a continuous function $S^1\to X$. So $f$ is trivially a loop... it depends on your definition as Mariano says – FShrike Nov 06 '22 at 23:06
  • @FShrike Usually it's defined as a map $l : I \to X$ with $l(0) = l(1)$. – Walker Nov 06 '22 at 23:06
  • @MarianoSuárez-Álvarez: wouldn't that not have $\pi_1(X,x_0)=0$? Sorry if this is a silly question, last time I thought about fundamental groups was in the 80s. – Martin Argerami Nov 06 '22 at 23:06
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    @SleepWalker As $S^1\cong[0,1]/(0\sim1)$, that is the same – FShrike Nov 06 '22 at 23:11
  • @MartinArgerami I think it would, but the original claim was that this should imply all other maps $S^1\to X$ are nullhomotopic, which can be false if $X$ isn't path connected and is false for $X=S^1\sqcup{x_0}$, just take the inclusion $S^1\hookrightarrow X$. – FShrike Nov 06 '22 at 23:26

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You need $X$ to be path-connected. Otherwise, Let's say $X=S^1\sqcup\{x_0\}$, i.e., $x_0\notin S^1$. Suppose there is no path from $x_0$ to any point of $S^1$. Then the only loop based at $x_0$ is the constant loop, so $\pi_1(X,x_0)=0$. With path-connectedness, we can talk about loops regardless of base points.

Let $f$ be any map from $S^1$ to $X$ (comment: in algebraic topology, we assume all maps are continuous; otherwise it is not interesting). Then $f$ maps $S^1$ to a loop on $X$, which is a trivial loop. it means that there is a path homotopy between $f(S^1)$ and the constant loop, by definition, it means that there is a path homotopy between $f(S^1)$ and a point. Thus $f$ is nullhomotopic.

Indeed, if $X$ is path-connected and has a trivial fundamental group, it is called simply-connected. This statement is equivalent to Proposition 1.6 in Hatcher's book.

And nullhomotopy of any map from $S^1$ to $X$ means that any loop on $X$ is a trivial loop, by the same statement, this is a characterization of simply-connectedness.

enihcamemit
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