# A confusion about the fact that contractible spaces are simply connected

Question 1: Greenberg's Algebraic topology has a proof that contractible spaces are simply connected. In the middle of the proof, the book makes use of the following fact without justifying it (probably because the author thinks its useful):

Fact 1: Let $p:\mathbb{I}\rightarrow X$ be a continuous function such that $p(0)=p(1)=x_0$ for some $x_0\in X$ . Let $c:\mathbb{I}\rightarrow X$ send every element of the interval to $x_0$. If $p,c$ are homotopic, then there exists a continuous function $F:\mathbb{I}\times \mathbb{I}\rightarrow X$ such that:

1) $F(s,0)=c(s)$ 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 $t\in \mathbb{I}$

Since the book is assuming this fact as obvious, I suppose it should be much easier to prove than to prove that contractible spaces are simply connected. I asked this here before and I got comments implying that fact 1 is almost as hard as proving the fact that "contractible spaces are simply connected". Well if this were true, is the proof of Greenberg's algebraic topology bad because it is assuming that fact 1 is obvious when proving that contractible spaces are simply connected while fact 1 is as hard as proving the original claim "contractible spaces are simply connected" ????

Important remark: I noticed that when I asked this question before, people were trying to prove the fact that "contractible spaces are simply connected" instead of answering the question. I know that it is tempting to answer this question without reading it carefully and just proving the fact that "contractible spaces are simply connected", please try to avoid this temptation :)

Question 2: Let's assume that fact 1 is true, since it's mentioned in the book. The following fact is also mentioned in the book. I will state it without proof:

Fact 2: Let $F:\mathbb{I}\times\mathbb{I}\rightarrow X$ be continuous such that $F(0,0)=F(0,1)=F(1,0)=F(1,1)$. Let $\gamma,\delta,\alpha,\beta :\mathbb{I}\rightarrow X$ be given by $\gamma (s)=F(s,0),\delta(s)=F(s,1),\alpha(t)=F(0,t),\beta(t)=F(1,t)$, then $\delta$ is homotopic to $\alpha^{-1} \gamma \beta$ $rel\{0,1\}$

Claim: Let $p:\mathbb{I}\rightarrow X$ be continuous such that $p(0)=p(1)=x_0$ for some $x_0\in X$. Let $c:\mathbb{I}\rightarrow X$ send every element of the interval to $x_0$. If $p,c$ are homotopic, then $p,c$ are homotopic rel$\{0,1\}$

Proof: Since $p,c$ are homotopic, therefore fact 1 can be used to show that there exists a homotopy $F$ between $p,c$ such that $F(0,t)=F(1,t), F(s,0)=c(s),F(s,1)=p(s)$ for all $t\in\mathbb{I}$. Set $\alpha(t)=F(0,t),\beta(t)=F(1,t)$. From fact 1, we know that $\alpha=\beta$. From Fact 2, we know that $p$ is homotopic to $\alpha^{-1} c \alpha$ rel $\{0,1\}$. Since $\alpha^{-1}c\alpha$ is homotopic to $\alpha^{-1}\alpha$ rel $\{0,1\}$ (because $c$ is constant map) , the last is homotopic to $c$ rel $\{0,1\}$. Therefore $p,c$ are homotopic rel $\{0,1\}$.

What is wrong with my argument in proving the above claim ?

I will explain here why I think my claim is not correct:

Consider the space $S$ formed as the identification space that results from gluing the the subspace $\mathbb{I}\times\{0\}\cup\{(0,1),(1,1)\}$ of $\mathbb{I}^2$ to one point. Let $F:\mathbb{I^2}\rightarrow S$ be the identification map. The space $S$ looks like the space resulting from gluing the three vertices of a triangular sheet of paper together (I already tried it using paper). Now note that $F:\mathbb{I}\times \mathbb{I}\rightarrow X$ is a homotopy between the paths $F|\mathbb{I}\times \{0\}$ and $F|\mathbb{I}\times \{1\}$. I fail to see that the paths: $F|\mathbb{I}\times \{0\},F|\mathbb{I}\times\{1\}$ are homotopic rel $\{0,1\}$ visually on the space $S$ that I created using paper. Thus, this is a potential counterexample to the claim .

If you have time, could you please construct the space $S$ from paper just like I did to understand me better ?

Question 3: I want to see a proof that contractible spaces are simply connected using the knowledge/definitions I have so far. I only know the definition of homotopic, homotopic rel $\{0,1\}$. Simply connected means having a trivial fundamental group. Contractible means that the identity map $1_X:X\rightarrow X$ is homotopic to some constant map $c:X\rightarrow X$

Finally thank you for wasting your time in answering my (probably) trivial questions

• I have had to delete my first answer, pretty picture or no, since it purports to prove false theorems. My current answer explains why. – Ryan Reich Oct 13 '13 at 5:54
• Isn't Fact 1 the same as your claim? – leo Oct 14 '13 at 23:15
• @leo: Actually, the claim is even more specific, if I understand what "rel $\{0,1\}$" is supposed to mean. Fact 1 furnishes a homotopy through loops; the claim furnishes a homotopy through loops with a fixed basepoint. – Ryan Reich Oct 15 '13 at 5:15

After some thought, I believe that neither Fact 1 nor the Claim are actually true as stated. Fact 2 is true, but does not imply Fact 1. The argument against Fact 1 is simple: let $X = S^1 = \{z \in \mathbb{C} \mid \lvert z \rvert = 1\}$ be the circle, let $p \colon [0,1] \to S^1$ be $p(t) = e^{2\pi i t}$, and let $c \colon [0,1] \to \{x_0\} = (1,0)$. Then $p$ is very much homotopic to $c$, say via $$H(s,t) = e^{2\pi i st};$$ we have $H(s,0) = c(s)$ and $H(s,1) = p(s)$. But $p$ is not homotopic to $c$ through loops, as Fact 1 claims. This is tricky to prove (how do you show that some homotopy doesn't exist?) and one would normally prove it with covering spaces, which show immediately that anything homotopic to $p$ has winding number $1$, while $c$ clearly has winding number $0$. I will simply leave it at that, since I don't think you can really do this with your current level of technology.

Note that any path is homotopic to any point along it, by a similar trick: just pick any $t_0 \in [0,1]$ and, since the interval is contractible, choose a continuous contraction $f(s,t)$ from the identity to $t_0$; for example, $$f(s,t) = t_0 + s(t - t_0)$$ (at $s = 0$ you get $t_0$, at $s = 1$ you get $t$, and since $0 \leq s \leq 1$, we have $f(s,t) \in [0,1]$; in fact, $f(s,t)$ is between $t$ and $t_0$ for all $s$). Then given a path $p \colon [0,1] \to X$, the map $F(s,t) = p(f(s,t))$ is a homotopy from $p$ to $p(t_0)$. As the other answer says, homotopy of paths without fixing endpoints is pretty meaningless.

Anyway, the Claim is an even more specific fact, that $p$ and $c$ must be homotopic through loops based at $x_0$. This is impossible for the same reason.

I am actually a bit puzzled where you got Fact 1 from. I have a copy of the book Algebraic Topology by Greenberg and Harper, presumably a revision, and the proof in question is covered entirely, and without unproven claims, by just Fact 2 (their Lemma 3.3). I believe you may have misinterpreted the key step in the following expanded version of the argument there, which does produce a homotopy with the properties given in Fact 1, but from stronger hypotheses.

Theorem: If $X$ is contractible, then it is simply connected.

Proof: Let $p \colon [0,1] \to X$ be any loop with $p(0) = p(1) = x_0$. Because of this, we can define a new continuous function $$q \colon S^1 \to X$$ where we consider $S^1 = [0,1]/(0 = 1)$. More explicitly, let $f \colon [0,1] \to S^1$ be the continuous map identifying $0$ with $1$, so we have $p = q \circ f$.

Suppose $X$ is contractible; then there is a homotopy $$H(x,t) \colon X \times [0,1] \to X$$ with $H(x,0) = x$ and $H(x,1) = x_0$ from the identity function to the constant function $x_0$. From this, we define a map $$H' \colon S^1 \times [0,1] \to X$$ by $H'(s,t) = H(q(s),t)$. This is a homotopy from $q$ to $x_0$, since \begin{align} H'(s,0) = H(q(s),0) = q(s) && H'(s,1) = H(q(s),1) = x_0 \end{align} It is continuous, being a composition of continuous functions. Define a further homotopy $$F(s,t) \colon [0,1]^2 \to X$$ by $F(s,t) = H'(f(s),t)$. It is continuous and a homotopy from $p$ to $x_0$, since \begin{align} F(s,0) = H'(f(s),0) = q(f(s)) = p(s) && F(s,1) = H'(f(s),1) = x_0. \end{align} $F$ has the property that \begin{align} F(0,t) &= H'(f(0),t) = H(q(f(0)),t) = H(p(0),t) = H(x_0,t) \\ &= H(p(1),t) = H(q(f(1)),t) = H'(f(1),t) = F(1,t). \end{align} That is, $F$ satisfies the criteria of Fact 1. (Note that we had to use the global homotopy $H$ to get $F$, not merely a homotopy from $p$ to $x_0$.) Defining $\alpha = \beta$ to be this common vertical-edge path, which is actually a loop at $x_0$ since $F(0,0) = F(0,1) = F(1,0) = F(1,1) = x_0$, we get from Fact 2 that $\alpha^{-1} p \alpha$ is homotopic to $x_0$ as loops based at $x_0$. That is, we have $$[\alpha]^{-1} [p] [\alpha] = [x_0] = 1$$ as homotopy classes in $\pi_1(X, x_0)$. Moving $[\alpha]$ to the other side, we get $[p] = [\alpha][\alpha]^{-1} = 1$, so $[p]$ is the trivial element of the fundamental group. Since $p$ is arbitrary, $X$ is simply connected. $\square$

Based on the manipulations in this proof, the best approximation to Fact 1 that I can come up with is:

Proposition: Let $X$ be a topological space and $f, g \colon X \to X$ two homotopic continuous maps. Then for any loop $p \colon [0,1] \to X$ based at $x_0 = p(0) = p(1)$, there is a homotopy $F \colon [0,1]^2 \to X$ from $f\circ p$ to $g\circ p$, both loops at $x_0$, such that for each $t \in [0,1]$, the path $s \mapsto F(s,t)$ is a loop based at $x_0$; i.e. $F(0,t) = F(1,t)$ for all $t$.

Proof: In short, make $p$ into a map $q$ from the circle and compose the assumed homotopy from $f$ to $g$ with $q$ in the first variable. Then, unwrapping the circle, you find that the resulting homotopy from $f \circ p$ to $g \circ p$ has the property claimed. The details are above. $\square$

• OK. So you think that fact 1 is incorrect. What can you say about this answer: math.stackexchange.com/questions/518848/… – Amr Oct 13 '13 at 9:48
• This answer is saying is that once you have a homotopy $F$ between a path $p$ with $p(0)=p(1)$ and a constant path that sends all elements of the interval to $p(0)$, you can turn it into a homotopy $G$ that satisfies the conditions of fact 1. Note that the proof of this answer does not use contractibility. – Amr Oct 13 '13 at 9:50
• Looks like Omar changed his answer in the last few hours. The previous answer was wrong since, as he noted, it was not a continuous homotopy. In fact, my example shows that it is impossible to prove Fact 1 without assuming contractibility (whereas it follows easily from Fact 2 when $X$ is contractible). – Ryan Reich Oct 13 '13 at 22:48
• Regarding the proof that contractible spaces are simply connected, I don't see how you go from $H(q(s),t)$ is a homotopy from $q$ to $x_0$. Also $F$ seems to be defined using $H'$. Could you please write the explicit defintion of $F$ from $H'$ ? Also if $H'$ is defined from $H$, could you please write its definition explicitly using $H$ ? – Amr Oct 14 '13 at 8:32

Question 1: I apologize again for trying to prove Fact 1. It is false, as Ryan already explained. A correct similar statement is that if $X$ is contractible, then for any path $p$ there exists a homotopy $F$ satisfying those conditions.

Question 2: Claims 2 is also false, and your proof of the claim is correct except for using Fact 1 which is false. By changing the first sentence of your proof you can turn it into a correct proof of the correct claim that if $X$ is contractible, for any loop $p$, $p$ and the constant map $c \equiv p(0) = p(1)$ are homotopic relative to endpoints.

Question 3: Here is how I think of the proof that contractible implies simply-connected. The whole thing boils down to showing your Fact 2: the definition of contractible you're using only gives homotopies that are not relative to endpoints, Fact 2 is the thing that converts those into homotopies rel endpoints.

I will use the following notation and facts:

1. By $p \sim q$ I'll mean that the two paths $p, q : [0,1] \to X$ are homotopic relative to the endpoints. (I won't use any symbol for homotopies that are not between paths or are not relative to endpoints.)

2. I'll use $q \cdot p$ to mean the concatenation of paths $p$ and $q$ (doing $p$ first and then $q$, say), and I'll need to use that this operation

1. respects homotopy in the sense that $p_1 \sim p_2 \implies q \cdot p_1 \sim q \cdot p_2$,
2. is homotopy associative, $p \cdot( q \cdot r) \sim (p \cdot q) \cdot r$, and
3. has homotopy units given by constant paths $c_x$, i.e., $p \cdot c_{p(0)} \sim p$, $c_{p(1)} \cdot p \sim p$.
4. has homotopy inverses given by traversing paths in reverse order, $p \cdot p^{\leftarrow} \sim c_{p(1)}$, $p^{\leftarrow} \cdot p \sim c_{p(0)}$.

I'm sure you've developed all of that if you know what the fundamental group is.

Now, let $X$ be contractible and choose a homotopy $H : X \times [0,1] \to X$ between the identity and a constant path $c_{x_0}$. Each point $x \in X$ then gets a chosen path $h_x$ connecting it to $x_0$, given by $h_x (t) = H(x,t)$. We can use your Fact 2 to produce some a homotopy relative endpoints for any path $p$: applying Fact 2 to the homotopy $(s,t) \mapsto H(p(s),t)$ we get a homotopy showing that $h_{p(1)} \cdot p \cdot h_{p(0)} ^ {\leftarrow} \sim c_{x_0}$. Using the above facts for the operation $\cdot$, we get that $p \sim h_{p(1)} ^\leftarrow \cdot h_{p(0)}$. This shows that all paths between $p(0)$ and $p(1)$ are homotopic to each other relative to endpoints because they are all homotopic to $h_{p(1)} ^\leftarrow \cdot h_{p(0)}$. This proves in particular that $X$ is simply-connected, because if $p(0)=p(1)$ we get $p \sim h_{p(1)} ^\leftarrow \cdot h_{p(0)} \sim c_{p(0)}$.

• Thanks Omar for your help. +1 – Amr Oct 16 '13 at 9:32

I do not know the book.

Let $p$ and $q$ be loops (closed paths) in $X$, i.e. $p,q\colon I\to X$, $p(0)=p(1)$, $q(0)=q(1)$. We can consider three concepts of homotopy between $p$ and $q$:

The first would be that there is a map $H\colon I\times I\to X$ such that $F(\bullet,0)=p$, $F(\bullet, 1)=q$.

The second would be that additionally $F(0,\bullet)=F(1,\bullet)$.

The third would be that $F(0,\bullet)$ and $F(1,\bullet)$ are even constant.

Now the Fact 1 that you have stated claims that the first condition implies the second (in the case that one of the paths is constant). This is certainly not true, and Ryan has pointed that out. Indeed, this first concept of homotopy is rather useless. According to it two loops are homotopic if and only if their images lie in the same path-component (exercise, but see at the end). So either the author was confused or you misread. What seems more likely is that the author meant that whenever he speaks of homotopies between paths he means those of the second kind.

The third kind is of course what you know as a homotopy relative to $\{0,1\}$ and what occurs in the definition of the fundamental group.

Ryan has also mentioned that this second kind of homotopy is a very natural concept if one thinks of a loop in $X$ as a map $\mathbb S^1\to X$ instead.

Now, generally homotopy relative base point (third kind) is stronger then free homotopy (second kind). However, and that is what you have shown with the help of Fact 2, a loop is freely homotopic to a constant loop if and only if it is homotopic relative $\{0,1\}$ to a constant loop.

You thought you had proved something even stronger because you used a wrong Fact 1, and that stronger statement is indeed false. Also your example for showing this is correct. Indeed, it suffices to identify $(0,1)$ and $(1,1)$.

Now to prove that contractible spaces are simply connected, it remains to show that in a contractible space every loop is freely homotopic (again, speaking of the second kind) to a constant loop. But this is easy, indeed just compose the loop with the homotopy showing that the space is contractible.

In case you feel that I have evaded you question one: If it was true then your proof (the one after Fact 2) would show that a space is simply connected if every loop in it allows a homotopy of the first kind to a constant loop. But every path $p$ does, just take $H(s,t)=p(ts)$. And there are non-simply connected spaces.

• Actually, your third-last paragraph is subject to my counterexample. There is really no relationship between free homotopy and based homotopy of loops unless you are using an auxiliary global homotopy to produce both of them. See my revised answer. – Ryan Reich Oct 14 '13 at 18:16
• I do not understand what you mean. What I try to say is that the notion of homotopy as in Fact 1 (and indeed with an arbitrary loop instead of the constant loop) is what you get if you identify the loops (closed paths) with maps from $\mathbb S^1$ and consider homotopies of maps from $\mathbb S^1$ without extra conditions. – Carsten S Oct 14 '13 at 18:47
• What I mean is that even with Fact 2, you can't show that a loop that is freely homotopic to a constant loop is also homotopic rel $\{0,1\}$ to a constant loop. Because I give an example of how the circle is freely homotopic to a constant loop, but has a nontrivial fundamental group, so can't be homotopic as a loop. – Ryan Reich Oct 14 '13 at 20:18
• I should clarify this in my answer. By freely homotopic I refer to a homotopy during which the path remains a loop at all times. The weaker concept is, as I wrote rather useless. – Carsten S Oct 14 '13 at 20:54
• @RyanReich, thank you for your criticism, I have reworked my answer. – Carsten S Oct 14 '13 at 22:32