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I've recently encountered a proof in my algebraic topology notes and the highlighted part is unclear to me. Why does $q(t_1, \tau_1) = q(t_2,\tau_2) \iff G(t_1, \tau_1) = G(t_2, \tau_2)$? I don't understand how the function q is deduced if we only know how $G$ behaves on the boundary of the unit square.

Any help/hint would be greatly appreciated.

Edit: In the lecture notes simply connectedness is defined as follows: A topological space is simply connected if it's both path-connected and also has the property that any continuous function mapping the boundary circle of a closed disc into the space can be extended continuously over the whole of the disk.

ben69234
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  • To me, this theorem is saying: "a path connected space is simply connected iff. it is simply connected". How do you define "simply connected"? For many, it is defined as [any loop is endpoint-homotopic to the constant loop] – FShrike Aug 03 '23 at 12:29
  • You should tell us which textbook it is and which definition of "simply connected " is used. – Paul Frost Aug 03 '23 at 12:50
  • @Paul Frost Apologies for the confusion. I've edited the question – ben69234 Aug 03 '23 at 13:32

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Usually "simply connected" is defined by the criterion formulated in Proposition 3.17, so your lecture notes use a non-standard approach.

To find $F$, one associates to $f$ the path $\gamma : I \to X$ and takes a homotopy $G : I \times I \to X$ relative to $\{0,1\}$ from $\gamma$ to the constant loop at $p$. Here $I = [0,1]$.

Let us consider the continuous map $q : I \times I \to D, q(t,\tau) = ((1-\tau)\cos(2\pi t) +\tau, (1-\tau)\sin(2\pi t))$.

The line segment $I \times \{\tau\}$ is mapped by $q$ onto the circle $C_\tau$ with center $(\tau,0)$ and radius $1 -\tau$ if $\tau < 1$ and to the one point set $C_1 = \{(1,0)\}$ if $\tau = 1$. Since $C_\tau \subset D$, we have in fact defined a map into $D$. Clearly we have $q(t,\tau) = (1,0)$ if and only if $\tau = 1$ or $t \in \{0,1\}$, i.e. $(t,\tau) \in R = I \times \{1\} \cup \{0,1\} \times I$. Hence $q$ restricts to a map $$q' : I \times I \setminus R \to D \setminus C_1 .$$ We show that $q'$ is a bijection.

  1. $q'$ is injective.

Let $z_i= (t_i,\tau_i)$, $i=1,2$, be two distinct points of $I \times I \setminus R$. If $\tau_1 \ne \tau_2$, then the $q'(z_i)$ lie in the disjoint sets $C_{\tau_i} \setminus C_1$. Thus $q'(z_1) \ne q'(z_2)$. If $\tau_1 = \tau_2 = \tau$, we must have $t_1 \ne t_2$. Since $t_1, t_2 \in (0,1) \subset I$, we get $(\cos(2\pi t_1), \sin(2\pi t_1)) \ne (\cos(2\pi t_2), \sin(2\pi t_2))$. Thus $q'(t_1,\tau_1) = q'(t_1,\tau) \ne q'(t_2,\tau) = q'(t_2,\tau_2)$.

  1. $q'$ is surjective.

It is intuively clear that $\bigcup_{\tau \in I} C_\tau = D$ which implies surjectivity. A formal proof is tedious and will be omitted here.

The above considerations have shown that $q$ is a surjection with the property that $q^{-1}(z) = R$ for $z = (0,1)$ and $q^{-1}(z)$ is a one point subset of $I \times I \setminus R$ for $z \in D \setminus C_1$.

Since $I \times I$ is compact and $D$ is Hausdorff, $q$ is a closed surjective map and therefore a quotient map.

By the above properties of $q$ we have $$q(t_1,\tau_1) = q(t_2,\tau_2) \implies G(t_1,\tau_1) = G(t_2,\tau_2) \tag{1}$$ because $G(R) = \{p\}$. This implies that there exists a (unique) function $F : D \to X$ such that $F \circ q = G$. The function $F$ is automatically continuous because $q$ is a quotient map.

Note that the reverse implication to $(1)$ is not needed (in general it is wrong).

Remark.

Also have a look at Characterizing simply connected spaces.

Paul Frost
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