I am studying for an exam in a little more advanced topology, and I want to understand why the mentioned map (i.e. forgetting the base points & hence allowing non-pointed homotopy) is an isomorphism (I know why it is onto, already).
In the lecture notes, the argument begins showing that for $ x_0 \in S^n $, $$ (S^n, \; x_0) \simeq \;((S^n, \;x_0) \vee (I, \;0), \;[1]) $$ where as usual, $I = [0,1]$ is the unit interval; $[1]$ denotes the equivalency class of $1$ in the wedge $(S^n, \;x_0) \vee (I, \;0) \overset{\text{(by def.)}} = (S^n\bigsqcup I)/\sim$, the quotient relation being generated by $x_0 \sim 0$.
This part I have understood.

However, I don't see how to use this to show that the map is injective. The notes say something about using that $S^n$ is simply connected. But how does this show that the base-point forgetting map (now using the indicated homotopy) $$ [(S^n \vee I, \;[1]), \;(S^n, x_0)]^\circ \longrightarrow [(S^n \vee I), S^n] $$ is injective?

The notes say something about using that $S^n$ is simply connected. I understand that any homotopy between morphisms $$ f,g:\;(S^n \vee I, \;[1]) \longrightarrow (S^n, \;x_0) $$ in the category of pairs of spaces induces a loop over $x_0$ by restricting that homotopy to $\{[1]\} \times I \cong I$, and that this probably plays a role, since any such loop is homotopic to a constant ($S^n$ simply connected).

If I'm not mistaken, I also understand, that, using the inclusions $$ \begin{align} &\iota_0:\; (S^n, \;x_0) \hookrightarrow (S^n \vee I, \;[0]) \\ &\iota_1:\; (I, \;0) \hookrightarrow (S^n \vee I, \;[0]) \\ \end{align} $$ allows us to write $f$ as a map induced by the two pointed maps $$ \begin{align} &f_0 = f \iota_0:\; (S^n, \; x_0) \longrightarrow (S^n, \;x_0) \\ &f_1 = f \iota_1:\; (I, \;0) \longrightarrow (S^n, \;x_0) \\ \end{align} $$ induced by the universal property of the sum in the category of pointed spaces (similarly, $g$ comes from $g_0$, $g_1$ then).

Now, given $f \simeq g$ by not necessarily pointed homotopy, how do I get a homotopy $$ H:\; (\; (S^n \vee I) \times I, \;\{[1]\} \times I \;) \longrightarrow (S^n, \;x_0) $$ ? If the given homotopy would split up nicely in accordance with how we split up $f,\; g$ into $f_0, \;g_0$ and $f_1, \;g_1$ respectively, then maybe we could use that (?). Sorry for the vague formulation in this paragraph, but I'm a little confused from thinking too much about this.

I already see criticism coming all ways for this, but please, do not give me 'Hatcher'-style answers to this. By this I mean: Please be more formal/rigorous/explicit (for instance, I like the book by/style of Tammo tam Dieck, there are also lecture notes by Lazarev on Algebraic Topology, and I believe I once read some notes by Emily Riehl. I also liked what I read from Spanier - all of these authors I know/remember to be 'more formal/rigorous/explicit').


migrated from mathoverflow.net Mar 30 '17 at 18:33

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  • $\begingroup$ By the way, I could not put a '\circ' between $f$ and $\iota_0$, $\iota_1$ inside the align environment, because the '\circ' would be converted into the string $aligned$ (or maybe, just $align$, I forgot). $\endgroup$ – polynomial_donut Mar 30 '17 at 14:07
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    $\begingroup$ This is a good question, but it belongs on MathStackexchange.com, rather than here. Have you tried using the Homotopy Extension Property (Chapter 0 of Hatcher)? $\endgroup$ – Mark Grant Mar 30 '17 at 15:19
  • $\begingroup$ As you seem to like tom Dieck's book, having a look at Chapter 6.2 (in particular Proposition 6.2.8) might please you. $\endgroup$ – archipelago Mar 30 '17 at 16:38
  • $\begingroup$ @archipelago Yes, but I'd like to avoid working through the prerequisites for 6.2.8 - we didn't touch much homotopy theory. Does the above approach work without 6.2.8. by any chance, with a more 'elementary' proof? $\endgroup$ – polynomial_donut Mar 30 '17 at 17:52
  • $\begingroup$ @MarkGrant Hatcher? No... just gave it a look in Dieck's book, though. Do you know if this gives a definitive/easy way toward continuing the above argument/ proving injectivity in another way? $\endgroup$ – polynomial_donut Mar 30 '17 at 18:12

Suppose $f,g:S^n\to S^n$ are homotopic maps. We want to show that they are based homotopic with respect to some base point $x_0\in S^n$. Of course, this doesn't make sense unless we assume that $f$ and $g$ are themselves based. But we are free to assume this, since every map is homotopic to a based map (this is the surjectiveness statement which you've already proved).

So, we have a homotopy $H:S^n\times I\to S^n$ from $f$ to $g$, where $f(x_0)=x_0$ and $g(x_0)=x_0$. If we look at what happens to the basepoint under this homotopy, we get a closed loop $H(x_0,-):\{x_0\}\times I\to S^n$. Since $S^n$ is simply connected, this loop is null-homotopic via a (based) homotopy $G:\{x_0\}\times I\times I\to S^n$.

Now we apply the homotopy extension property for the pair $\{x_0\}\times I\subseteq S^n\times I$ (a closed cofibration). The map $$H\cup G:S^n\times I\times \{0\}\cup \{x_0\}\times I\times I\to S^n$$ extends to a homotopy $F:S^n\times I\times I\to S^n$. This is homotopy from our original homotopy $H=F|_{S^n\times I\times\{0\}}$ to a based homotopy $H'=F|_{S^n\times I\times\{1\}}:S^n\times I\to S^n$.

  • $\begingroup$ Hmm I wonder why, but we didn't really do the HEP in Topology I, and not (yet/up to where I am in my studies of the lecture) in our Topology II class. Hopefully I'll get to catch up on that soon enough. I suppose this is the only sensible way to prove this using that the S^n is simply connected (?). $\endgroup$ – polynomial_donut Apr 13 '17 at 3:11
  • $\begingroup$ @polynomial_donut: Yes, any proof must use simple-connectedness of $S^n$, because the result is not true for a general space $X$. In general, the unbased homotopy set $[X,X]$ is a quotient of the based homotopy set $[X,X]^\circ$ under an action of $\pi_1(X)$. $\endgroup$ – Mark Grant Apr 13 '17 at 6:15
  • $\begingroup$ @Mark_Grant Thank you for answering. My question was if there was another way to prove this (possibly using simple-connetedness) without the HEP... (I guess/suppose not/ or it wouldn't make much sense, but I'm curious, because we hadn't had the HEP in our lecture - it's very well possible, that the professor simply ignored that fact with his remark). $\endgroup$ – polynomial_donut Apr 13 '17 at 12:56
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    $\begingroup$ @polynomial_donut: Perhaps the full force of the HEP is not required here. It's probably enough to note that the space $S^n\times I\times\{0\}\cup\{x_0\}\times I\times I$, which is a cylinder with a square attached, is a retract of $S^n\times I\times I$. This gives the required extension. The HEP seems the right way to understand it though. $\endgroup$ – Mark Grant Apr 13 '17 at 13:27

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