I am having trouble justifying exactly why a loop in $X$ is the same thing as a map $f\colon S^1\to X$. A loop is a map $g\colon \lbrack 0,1\rbrack\to X$ with $g(0)=g(1)$. How can I use the fact that $S^1$ is homeomorphic to the quotient $\lbrack 0,1\rbrack/\sim$, where $\sim$ is the relation $0\sim 1$, to show that any loop is really $f\colon S^1\to X$?
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1Are you familiar with the universal property of quotient maps? – Leon Sot Dec 12 '16 at 03:55
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No I haven't seen that before. – merkinator Dec 12 '16 at 04:53
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The universal property of equivalence classes (this should be just a matter of Linear Algebra, first semester):
If $f:X\to Y$ is a map with an equivalence relation $R$ on $X$ and for every $x,x'\in X$ with $x\sim x'$ we have $f(x)=f(x')$, then there exists a unique map $\bar{f}:X/\sim\to Y$ with $\bar{f}\circ \pi=f$.
This means for our case, we have a commutative diagramm $$\matrix{ [0,1] & \overset{f}{\longrightarrow} & X \\ \downarrow \pi & \nearrow \exists !h& \\ \frac{[0,1]}{0\sim 1}\cong S^1 & & }$$ Since the natural projection $\pi$ and $f$ are continuous, $h$ is continuous (use the quotient topology of $[0,1]/0\sim1$). For the proof of the universal property of the quotient see Proof of the universal property of the quotient topology.
So we have $f(x)=(h\circ \pi)(x)\quad\forall x\in[0,1].$
Daniel Bernoulli
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