# Question on showing a bijection between $\pi_1(X,x_0)$ and $[S^1, X]$ when X is path connected.

I am trying to do this question taken from Hatchers algebraic topology and I am struggling to understand the notation and the concepts.

As far as I know $\pi_1(X,x_0)$ is the set of end point preserving homotopy classes of loops in X based at $x_0$ and a loop is just a path $f:I \rightarrow X$ with $f(0)=f(1)$. The question says we can regard $\pi_1(X, x_0)$ as the set of basepoint preserving homotopy classes of maps $(S^1, s_0) \rightarrow (X, x_0)$.

This confuses me, does it mean the set of basepoint preserving homotopy class on maps $g:S^1 \rightarrow X$ which map loops in $S^1$ based at $s_0$ to loops in X based at $x_0$?

Then how can $\pi_1(X,x_0)$ be regarded as this set?

If someone could explain this it would be really useful

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A based map $(S^1,s_0) \to (X,x_0)$ is a map $f: S^1 \to X$ with $f(s_0) = x_0$. (This is what a based map is for any pair of pointed spaces.)

Your notion of loop is the same as the notion of a based map from $S^1$. Note that $I/(0 \sim 1) \cong S^1$ as a topological space. By the definition of a quotient map, if you have a continuous map $f: I \to X$ such that $f(0)=f(1)=x_0$, this factors through the above quotient map, giving a map $f': I/(0 \sim 1) \to X$, with $f'([0]) = x_0$.

Conversely, given a based map $f: (S^1, s_0) \to (X,x_0)$ you can obtain a map $f': I \to X$, with $f'(0)=f'(1)=x_0$, essentially by undoing the above process.

Once you understand why these two notions of loop are actually (in the above sense) the same, it should not be difficult to go one step further and see why you can consider $\pi_1$ as (based) homotopy classes of (based) maps from $S^1$.

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Ok thanks, that explains a lot. So for the first bit, if we let $[\alpha]\in [S^1,X] \cong [I/(0\sim 1),X]$ we know that $\alpha$ maps $S^1$ to some loop in X with some basepoint x. There exists some path $h$ connecting $x$ to $x_0$ and the loop $\beta$ formed from concatenating $h$, $\alpha$ and $\bar{h}$ is homotopic (not basepoint preserving but it doesn't matter) to $\alpha$ but $[\beta]\in \pi_1(X,x_0)$ and we have that $\phi([\beta])=[\alpha]$. Does that make any sense? I'm unsure whether the concatenation works because $\alpha : S^1 \rightarrow X$ –  Tom Mar 22 at 18:24
@Tom: I don't want to give too many details, since this is a good exercise to fill in the details of. You now have a map $\pi_1(X) \to [S^1,X]$ (take a loop, turn it into a map from $S^1$). You have to first show this is well-defined. That is, that if the path $\gamma$ is (basepoint-preserving!) homotopic to $\alpha$, then $\gamma$ maps to the same element of $[S^1,X]$ as $\alpha$. You should be able to see why this map is onto. –  Mike Miller Mar 22 at 19:28
Now, as to why two conjugate elements of $\pi_1(X)$ map to the same element of $[S^1,X]$ - try drawing some pictures for ideas. A good test case is $X = S^1 \vee S^1$, which has fundamental group free on two generators; try to see why $g_1g_2g_1^{-1}$ is homotopic to $g_2$ as a map from $S^1$. (Draw a picture! In the general case, you'll use that $X$ is path connected.) –  Mike Miller Mar 22 at 19:30
Ok, I've been doing some thinking, sorry for being a bit slow I find this stuff confusing. So let $\gamma \sim \alpha$ (bp preserving) be loops based at $x_0$ in $X$. $\gamma,\,\alpha$ can be viewed as based maps from $(S^1, s_0)$ to $(X, x_0)$. Under the map $\phi$ we remove the restriction on our base points $\phi([\gamma]), \phi([\alpha])$ are now the homotopy (non-bp) classes, however since any two maps which are bp-preserving homotopic are certainly non-bp preserving homotopic we have a well defined map Is this correct reasoning? Mainly the bit in bold –  Tom Mar 23 at 10:19
Your logic is almost correct, but not quite. The key is that a homotopy looks a little different now. A homotopy between $\alpha$ and $\gamma$ is a map $f: I \times I \to X$ with $f(t,0) = \alpha(t)$, $f(t,1) = \gamma(t)$, and $f(0,s) = f(1,s) = x_0$. The point is that this induces a homotopy between $\varphi(\alpha)$ and $\varphi(\gamma)$; simply glue the 'sides' of $I \times I$ together to get a map from $I \times I /((0,s) \sim (1,s))$. This is precisely a map from $S^1 \times I$ - same logic as before. And a map from $S^1 \times I$ that restricts to the appropriate maps is a homotopy. –  Mike Miller Mar 23 at 10:21