In greeberg's algebraic topology, the following fact is used in the proof that contractible spaces are simply connected without justification:

Let $p:\mathbb{I}\rightarrow X$ be a continuous function such that $p(0)=p(1)$ and $X$ is contractible. Then there exists a continuous function $F:\mathbb{I}\times \mathbb{I}\rightarrow X$ such that:

1) $F(s,0)=p(0)$ for all $s\in \mathbb{I}$

2) $F(s,1)=p(s)$ for all $s\in \mathbb{I}$

3) $F(0,t)=F(1,t)$ for all $t\in \mathbb{I}$

I agree that there exists a map $F$ such that the first 2 conditions are satisfied, but what about the third ? Is there an obvious way to see that that the three conditions can be satisfied. As I mentioned the book uses this fact without justification.

Thank you

Reminder: All answers posted so far are proving the fact that contractible spaces are simply connected. Note that my original question (found above) asks for a justification of the fact that the book stated without justification. I don't think it s obvious.

  • $\begingroup$ About your "Reminder": there is only a tiny tiny difference between the unjustified fact and a proof of simply connectedness, which is probably why it feels as if the answers are giving you a full proof "instead" of just proving that fact. $\endgroup$ Oct 8 '13 at 15:52
  • $\begingroup$ @OmarAntolin-Camarena Really ?? This fact was used without justification in the middle of an argument to prove that contractible spaces are simply connected and I was feeling stupid for not seeing why a fact that is used as obvious is really obvious.... Anyways thanks for your time $\endgroup$
    – Amr
    Oct 8 '13 at 16:49
  • $\begingroup$ I don't see the difference at all. The result above, and the fact that $X$ contractible implies $X$ simply connected are equivalent, provided "contractible" means that there is a deformation retraction onto any point, in particular $p(0)$. If anything, though, the above result is stronger than contractible implies simply connected. $\endgroup$ Oct 10 '13 at 5:20

I apologize for the previous false starts, here's a simple way to use contractibility to construct an $F$ satisfying (1), (2) and (3). Notice, in particular, that such an $F$ does not need to have $F(0,t)$ constant, just needs to have it equal to $F(1,t)$ for each $t$.

OK, let $H : X \times [0,1] \to X$ be a homotopy of the identity on $X$ to the constant map with value $p(0)$, that is, let $H(x,0) = p(0)$, $H(x,1) = x$, for all $x$ (note that we do not assume that $H$ fixes $p(0)$ for all $t$, as that cannot always be arranged). Now define $F(s,t) = H(p(s),t)$. We have that

  • $F(s,0) = H(p(s),0) = p(0)$,
  • $F(s,1) = H(p(s),1) = p(s)$, and
  • $F(0,t) = H(p(0),t) = H(p(1),t) = F(1,t)$.
  • $\begingroup$ Dear Omar it thanks for your help. It was difficult for me to follow your "visual" argument> It seems to me you did not handle the rel {0,1} issue (not sure as I did not understand your answer). $\endgroup$
    – Amr
    Oct 8 '13 at 13:44
  • $\begingroup$ Another note: My question actually does not ask for a full proof that contractible spaces are simply connected. I just need a justfication for a fact that the book used in the middle of a proof without justifying it. However if you can also give me an argument for showing that contractible spaces are simply connected its ok. I am just reminding you what my question originally asked $\endgroup$
    – Amr
    Oct 8 '13 at 13:46
  • $\begingroup$ The $G$ I described is (almost*) the $F$ you want satisfying (1) through (3). The construction I gave does take care of the rel {0,1}: making the homotopy rel {0,1} just means that each $G_t$ is aloop starting and ending at $p(0)=p(1)$. $\endgroup$ Oct 8 '13 at 15:50
  • $\begingroup$ I've now completely rewritten the answer so it address only what you wanted (I apologize for answering slightly more than was asked before), and so it is not "visual" anymore, @Amr. $\endgroup$ Oct 8 '13 at 16:29
  • $\begingroup$ Thanks for your time. I will read your answer after a while. $\endgroup$
    – Amr
    Oct 8 '13 at 16:45

I think the question is asking about proving that for a contractible space $X$ a map $(S^1,1) \to (X,x)$ is null homotopic relative to the base point $1$. Of course a contracting homotopy of $X$ may move the base point $x$.

It is often easier to understand these matters from a more general viewpoint. The key property of $(S^1,1)$ is that it is well pointed, i.e. has the homotopy extension property. So a relevant lemma is essentially as follows, and is 7.2.11 of Topology and Groupoids. Recall an inessential map is one homotopic to a constant map.

Let $f:Y \to X$ be an inessential map, and suppose $y$ is well pointed in $Y$. Then $f$ is inessential rel $y$.

The feature of the proof is that that the null homotopy of $f$ defines a path $\alpha$ in $X$ from $x=f(y)$ to $x'$ say. Because $(Y,y)$ is well pointed, $\alpha$ defines a bijection $\alpha_*: [(Y,y), (X,x)] \to [(Y,y), (X,x')]$, where these are homotopy classes relative to the base point. Let $g: Y \to X$ be the constant map with value $x$. Then $\alpha_*[f]= \alpha_*[g]$ and so $[f]=[g]$.

These operations generalise the operations of the fundamental group on higher homotopy groups. In moving from $(S^n,1)$ to $(X,A)$ and looking at the fact that a homotopy equivalence of spaces induces an isomorphism of homotopy groups, I discovered a gluing theorem for homotopy equivalences (Section 7.4 of the above book.)

  • $\begingroup$ So, this is a more general fact and there genuinely seems to be a gap in the proof in Greenberg's book, as the OP claims? $\endgroup$
    – user96815
    Oct 13 '13 at 19:06
  • $\begingroup$ Dear Dr. Brown I already understood the confusion that happened regarding the fact that "contractible spaces are simply connected". I have something else to ask about. I read chapters: 4-6,8 of your book topology and groupoids. Now I am reading greenberg's AT. I found that it's neater to state theorems using groupoids, I also liked the theorem van-kampen theorem of chapter 6. When I first started reading the book, I had the impression that the fundamental groupoid functor $\pi:Top\rightarrow Grpd$ would be better at differentiating path connected spaces than (to be continued) $\endgroup$
    – Amr
    Oct 14 '13 at 18:18
  • $\begingroup$ the combined use of of the fundamental group and the forgetful functor $F:Top\rightarrow Set$. However, a negative answer to this question math.stackexchange.com/questions/419939/… showed that my inital impression was false $\endgroup$
    – Amr
    Oct 14 '13 at 18:20
  • 1
    $\begingroup$ My question is: Is the advantage of using groupoids instead of the fundamental group and $F:Top\rightarrow Set$ only in getting neater theorems or there is something else ? Of course I appreciate the fact that we're getting theorems, but I am just asking if this is the only advantage. Thank you. $\endgroup$
    – Amr
    Oct 14 '13 at 18:22
  • $\begingroup$ @Amr: The original aim was to get a theorem which also calculated the fundamental group of the circle, which is, after all, THE basic example in algebraic topology. Notice also that Section 8.4 is used in Section 9.2. The further advantage to me was that it eventually suggested the possibility of higher homotopy van Kampen Theorems, and these allow new calculations in homotopy theory. I also feel maths progresses by getting "neater theorems"! Trying to find why something is true needs the "right" structure. See also Grothendieck's comments on my web page for the book. $\endgroup$ Nov 5 '13 at 14:52

A space $X$ is simply connected if and only if every continuous map $p: S^1\to X$ extends to a continuous map $P: D^2\to X$. Now, if you have a contraction $F: X\times I\to X$ of $X$, then simply take $$ P(r e^{it})= F(p(e^{it}), r). $$ Here I am regarding $D^2$ as the unit disk in the complex plane.

  • $\begingroup$ The defintion of contractin, I am using is that the identity map $1_X:X\rightarrow X$ is homotopic to the constant map $r:X\rightarrow X$ $\endgroup$
    – Amr
    Oct 8 '13 at 12:31
  • 1
    $\begingroup$ A fact that I also proved is that a space is contractible iff every two maps $f,g\in Top(Y,X)$ are homotopic $\endgroup$
    – Amr
    Oct 8 '13 at 12:32
  • $\begingroup$ @Amr: My map $F$ is the homotopy from constant map to the identity. Which part of the proof did you find unclear? $\endgroup$ Oct 8 '13 at 13:38
  • $\begingroup$ I am sorry I meant I am using a different defintion for simple connectedness. The one I am using is about the trivial fundamental group $\endgroup$
    – Amr
    Oct 8 '13 at 13:40
  • $\begingroup$ @Amr: The two definitions are equivalent. $\endgroup$ Oct 8 '13 at 13:43

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