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Contractible spaces have trivial fundamental groups.

The definition of contractible spaces which I am following is as follows $:$

Definition $:$ A space $X$ is said to be contractible if it is homotopy equivalent to a point (as a non-based space).

If $X$ were based homotopic to a point then contractibility of $X$ would imply that $X$ has trivial fundamental group. But what will happen if $X$ is not based homotopic to a point? Can I still able to conclude the same?

ACB
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2 Answers2

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If $f \colon X \to Y$ is a homotopy equivalence of spaces, it induces an isomorphism of fundamental groups

$$f_\ast \colon \pi_1(X, x_0) \to \pi_1(Y,f(x_0)).$$

If $Y= \ast$ and then this implies $\pi_1(X,x_0) = \ast$.

As you say, in proving this satement there is something to be said about basepoints. Namely, given $x_0 \in X$ and a homotopy inverse $g$ of $f$, even if $gf \simeq 1$ we need not have $gf(x_0) = x_0$. Hence $(fg)_\ast$ is not the identity on $\pi_1(X,x_0)$, it is not even an endomorphism (it is a map $\pi_1(X,x_0) \to \pi_1(X,gf(x_0))$).

Nonetheless, for path-connected spaces changing of basepoint via a fixed path induces an isomorphism at the $\pi_1$-level.

Details can be found in Hatcher's book: the fact that homotopy equivalences induce isomorphisms on $\pi_1$ is Proposition 1.18 and the basepoint subtleties are dealt with in Lemma 1.19.

qualcuno
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A one point space $P$ is characterized by the property that there exists only one map $r^Y_P : Y \to P$, where $Y$ is arbitrary.

$X$ being contractible can be defined as being homotopy equivalent to a one point space $P = \{*\}$. This means that there exist maps $f : P \to X$ and $r : X \to P$ such that $r \circ f \simeq id_P$ and $f \circ r \simeq id_X$. Clearly $r = r^X_P$ and $r \circ f = r^P_P = id_Y$. Thus contractible means that $f \circ r^X_P \simeq id_X$ for some $f : P \to X$. This is equivalent to the existence of a homotopy $H : X \times I \to X$ such that $H_0 = id_X$ and $H_1$ is a constant map (here $H_t : X \to X, H_t(x) = H(x,t)$ for $t \in I$). This property is usually taken as the definition of contractible.

In fact, if $f \circ r^X_P \simeq id_X$, then there is a homotopy $H : X \times I \to X$ such that $H_0 = id_X$ and $H_1 = f \circ r^X_P$ which is a constant map. Conversely, given a homotopy $H$ as above, there exists $x_0 \in X$ such that $H_1(x) = x_0$ for all $x$. Now take $f : P \to X, f(*) = x_0$.

The fundamental group $\pi_1(X,x)$ can be represented as the set of pointed homotopy classes of pointed maps $u : (S^1,s_0) \to (X,x)$. If $X$ is contractible, then $u = id_X \circ u \simeq c \circ u$, where $c : X \to X$ is some constant map. Hence there exists a homotopy $G: S^1 \times I \to X$ such that $G_0 = u$ and $G_1$ is constant. The map $p : S^1 \times I \to D^2, p(x,t) = (1-t)x$, is a quotient map identifying $S^1 \times \{1\}$ to a point. Thus $G$ induces a map $\bar G : D^2 \to X$ such that $\bar G \circ p = G$. We have $\bar G(z) = G(z,0) =u(z)$ for $z \in S^1$. Define $H : S^1 \times I \to X, H(z,t) = \bar G((1-t)z +ts_0)$. This is a well-defined continuous map (note that $\lVert (1-t)z +t \rVert \le (1-t)\lVert z \rVert + t \lVert s_0 \rVert = 1 - t + t = 1$). We have $H(z,0) = \bar G(z) = u(z), H(z,1) = \bar G(s_0) = u(s_0) = x$ and $H(s_0,t) = \bar G(s_0) = u(s_0) = x$. This shows that $u$ is pointed homotopic to the constant loop based at $x$.

See also Characterizing simply connected spaces.

Paul Frost
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  • Paul, a question. For $G$ to be continuous shouldn't the homotopy between $u$ and the constant be free? – Leandro Caniglia Jun 19 '21 at 20:30
  • @LeandroCaniglia Yes, it is a free homotopy. The only requirement is $G_0 = u$ and $G_1$ constant. The intermediate $G_t$ are allowed to be arbitrary maps. – Paul Frost Jun 19 '21 at 22:31
  • Maybe I wasn't clear. For $G$ to be continuous we need a homotopy $L$ satisfying $L(0,\tau) = L(1,\tau)$ for all $\tau\in[0,1]$. Don't we? – Leandro Caniglia Jun 19 '21 at 22:59
  • @LeandroCaniglia Yes. This is what I say in my answer. I consider pointed maps $u:(S^1,s_0) \to (X,x)$ which are freely homotopic. This is equivalent to considering closed paths and homotopies as in your last comment. See also https://math.stackexchange.com/q/4044399. – Paul Frost Jun 19 '21 at 23:44
  • Let me rephrase my question: Why the contractility of $X$ implies the existence of $G$? Is your proof self-contained or are you using another result? – Leandro Caniglia Jun 20 '21 at 11:38
  • @LeandroCaniglia I use the fact that the fundamental group $\pi_1(X,x)$ can be represented as the set of pointed homotopy classes of pointed maps $u : (S^1,s_0) \to (X,x)$. Most frequently the fundamental group is defined as the set of closed paths based at $x_0$ modulo homotopy of such paths. That both descriptions are equivalent is easy to see; just note that $[0,1]/{0,1} \approx S^1$. See also the linked question Characterizing simply connected spaces. – Paul Frost Jun 20 '21 at 14:19
  • I've finally understood the link I was missing: If $X$ is contractible, any map $Z\to X$ is homotopic to a constant. You are applying this fact to the case $Z=\mathbb S^1$. Thanks for your patience. – Leandro Caniglia Jun 21 '21 at 18:15