a simple arguement that $S^\infty$ is simply connected.

Question

Think of $$S^1 \subset S^2\subset S^3 \dots S^k \subset S^{k+1} \dots$$

Then consider $$S^{\infty}=S^1 \cup S^2 \cup S^3 \dots$$
How do I show that $S^\infty$ is simply connected.

I think if $\gamma$ is a path in $S^{\infty}$ then it is compact. Hence it is contained in some $S^n$ and then we can find a homotopy of $\gamma$ to a point in $S^n$ and this will give us the homotopy of $\gamma$ with a constant path. This will prove $S^{\infty}$ is simply connected.

By Homotopy groups of simplicial sets commute with filtered colimits, $\pi_1(\varinjlim S^n)=\varinjlim\pi_1(S^n)=\varinjlim\pi_1(S^{n+1})=\varinjlim1=1.$ — Anne Bauval, Dec 16 '22 at 22:38
The title of your question is about $S^n$ being simply connected, but the content of your question is $S^{\infty}$ being simply connected (and it looks like you're willing to assume already that $S^n$ is simply connected). — Thorgott, Dec 16 '22 at 22:39
See also https://ncatlab.org/nlab/show/infinite-dimensional+sphere — Anne Bauval, Dec 16 '22 at 22:43
@Thorgott, Sorry. I meant I wanted to show $S^\infty$ is simply connected. — permutation_matrix, Dec 17 '22 at 04:13
$S^{\infty}$ is contractible, and so it is simply connected. See https://math.stackexchange.com/questions/63349/is-s-infty-contractible-does-it-have-cw-structure, https://mathoverflow.net/questions/198/how-do-you-show-that-s-infty-is-contractible, https://math.stackexchange.com/questions/282268/unit-sphere-in-mathbbr-infty-is-contractible, https://math.stackexchange.com/questions/4135280/understanding-the-proof-that-s-infty-is-contractible, etc. — John Palmieri, Dec 17 '22 at 04:15
@JohnPalmieri, I know that. I wanted to know if this method works with a modification — permutation_matrix, Dec 17 '22 at 04:33
Then see https://math.stackexchange.com/questions/3679488/homotopy-groups-of-s-infty/3679501#3679501. — John Palmieri, Dec 17 '22 at 05:16

a simple arguement that $S^\infty$ is simply connected.

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