In Hatcher 1.1.6

We can regard $\pi_1(X,x_0)$ as the set of basepoint-preserving homotopy classes of maps $(S_1,s_0) \to (X,x_0)$. Let $[S_1,X]$ be the set of homotopy classes of maps $S_1 \to X$, with no conditions on basepoints. Thus there is a natural map $Φ: \pi_1(X,x_0) \to [S_1,X]$ obtained by ignoring basepoints.

I don't quite understand what is mean by "with no condition on base points"(What is mean by "base point" of $S_1$?) and "ignoring basepoints". And whether $[S_1,X]$ is defined to be a set of continuous map or all the maps. What is a formal description of "ignoring basepoints"?

If some one could give an example that given a map in $\pi_1(X,x_0)$, how to get maps in $[S_1,X]$ in the way the author describes. I would very appreciate.


Map here does mean continuous function. Here $[S^1,X]$ is the set of equivalence classes of continuous maps from $S^1$ to $X$, where two maps are equivalent if they are homotopic. On the other hand $\pi_1(X,x_0)$ is the set of equivalence classes of "basepoint-preserving" continuous maps $S^1\to X$ where two maps are equivalent by a "basepoint-preserving" homotopy. So this is a smaller class of maps factored by a finer equivalence, so we get a map from $\pi_1(X,x_0)$ to $[S^1,X]$.

In the text Hatcher defines $\pi_1(X,x_0)$ in a different way, as the set of continuous maps from $I$ to $X$ with $f(0)=f(1)=x_0$ factored out by a suitable equivalence relation. But a continuous map from $I$ with $f(0)=f(1)$ is the same as a map from $S$ where $S$ is the quotient space of $I$ got by identifying $0$ and $1$. But this $S$ is homeomorphic to $S^1$. If we let $s_0$ be the point on $S=S^1$ to which $0$ and $1\in I$ are identified, then continuous maps from from $I$ to $X$ with $f(0)=f(1)=x_0$ are the same as continuous maps from $S^1$ to $X$ with $f(s_0)=x_0$, that is "basepoint-preserving" maps. Also homotopies need to be "basepoint-preserving" too...

  • $\begingroup$ Thanks for the explaination. But I cannot imagine a non-basepoint-preserving map from $S^1\to X$ yet. I think all these maps in $[S^1,X]$ should be basepoint-preserving since otherwise the function is not well- defined at some point, but it seems that such map exists. Could you tell me an example of a non-base point-preserving continous map from $S^1$ to $X$? $\endgroup$ – user464003 Aug 3 '17 at 5:15
  • $\begingroup$ I have edited the title of the post and hence removed your initial comment. Regards, $\endgroup$ – Pedro Tamaroff Aug 3 '17 at 5:15
  • $\begingroup$ @user464003 Let $S^1$ have base point $1$ and consider a nontrivial rotation of itself. $\endgroup$ – Pedro Tamaroff Aug 3 '17 at 5:16
  • $\begingroup$ @user464003 Take a point $x_1\ne x_0$ and define $f:S^1\to X$ by $f(s)=x_1$ for all $s$. $\endgroup$ – Angina Seng Aug 3 '17 at 5:18
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    $\begingroup$ @LordSharktheUnknown Just an idle thought, but if someone is struggling to understand the connection between $\pi_1(X, x_0)$ and $[S^1,X]$, has just begun on a book like Hatcher and has trouble seing what a phrase like "ignoring basepoints" means, then I would say that the odds that they are intricately familiar with the notion of forgetful functors, on the level that it actually aids them in understanding those other things, is quite low. It would be cool if I was wrong, though. $\endgroup$ – Arthur Aug 3 '17 at 9:29

An element $\gamma\in\pi_1(X,x_0)$ is an equivalence class of functions. Take one function from $\gamma$. It is (or can, at least, be seen as) a function from $S^1$ to $X$ where a specific point of $S^1$ is sent to $x_0$ (i.e. base point preserving).

However, at the same time it is also just a function from $S^1$ to $X$ (we just "forget" that the function has this base point preserving property). Therefore it is part of some equivalence class in $[S^1,X]$. The point of Hatcher is that we get the same equivalence class of $[S^1,X]$ no matter which representative of $\gamma$ we choose.


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