4
$\begingroup$

Unfortunately, this becomes a very general post: I have some questions concerning the homotopy invariance of homotopy groups. I start from what should be clear:

  1. If $f,g:(X,x_0)\to (Y,y_0)$ are based homotopic, then $\pi_k(f)=\pi_k(g)$, which is immediately clear from the definition of the homotopy groups. In particular, we get $\pi_k(X,x_0)\cong\pi_k(Y,y_0)$ for based homotopy equivalent spaces and $\pi_k(X,x_0)=0$ for based contractible spaces.

  2. There are maps $f,g:(X,x_0)\to (Y,y_0)$ which are freely homotopic, but not based homotopic: Consider e. g. $f,g:(\mathbb{S}^1,\infty)\to (\mathbb{S}^1\vee\mathbb{S}^1,\infty)$, $f$ corresponds to $x$ and $g$ corresponds to the concatenation $y\star x\star y^{-1}$ where $x$ and $y$ denote the two canonical loops generating $\pi_1(\mathbb{S}^1\vee\mathbb{S}^1,\infty)$. This means the map $[(\mathbb{S}^1,\infty),(\mathbb{S}^1\vee\mathbb{S}^1,\infty)] \to [\mathbb{S}^1,\mathbb{S}^1\vee\mathbb{S}^1]$ is not injective. It is clear that $\pi_1(f)\ne\pi_1(g)$ although $f$ and $g$ are freely homotopic. (true?)

  3. Even worse, there are based spaces, which are (freely) contractible, but not based contractible: Let $(H,0)$ be the Hawaian earring and consider the cone $(\mathscr{C}H,[0,0])$ (i. e. we choose as a basepoint not the cone tip). Then as a cone, $\mathscr{C}H$ is contractible, but every contraction moves the basepoint. (true? The “problem” may be that $0\hookrightarrow H$ is not a cofibration?)

  4. If $(X,x_0)$ is a good pair (i. e. the inclusion $x_0\hookrightarrow X$ is a cofibration), then every path $\gamma:I\to Y$ ending in $y_0$ gives rise to a map $\gamma_\#:[(X,x_0),(Y,\gamma(0))]\to [(X,x_0),(Y,y_0)]$ by extending $f\vee \gamma:(X,x_0)\vee (I,0)\to Y$ to a map $H:X\times I\to Y$ and defining $\gamma_\# f(x) := H(x,1)$.

    • If $Y$ is connected, every map $f:X\to Y$ is (freely) homotopic to a based map: Choose a path $\gamma:I\to Y$ from $f(x_0)$ to $y_0$, then $H$ from above is a homotopy $f$ to the based map $\gamma_\# f$.

    • The map $(\gamma,f)\mapsto \gamma_\# f$ gives rise to an action of $\pi_1(Y,y_0)$ on $[(X,x_0),(Y,y_0)]$ and it can be easily seen that $[X,Y]=[(X,x_0),(Y,y_0)]/\pi_1(Y,y_0)$.

My actual questions are:

  1. The statement “If $f,g:(X,x_0)\to (Y,y_0)$ are homotopic, then $\pi_k(f)=\pi_k(g)$” seems to be wrong, already for the fundamental group, even for well-pointed spaces. What is the “next best” statement which is true?

  2. Do we at least know something like “If $(X,x_0)$ and $(Y,y_0)$ are (freely) homotopy equivalent, then $\pi_k(X,x_0)\cong \pi_k(Y,y_0)$.”, which should be the “easy” direction of the Whitehead theorem? Maybe we need some well-pointedness condition?

  3. If (2) fails, do we at least know something like “If $X$ is contractible, then $\pi_k(X,x_0)=0$ for each $k\ge 0$ and $x_0\in X$?”

I apologize for the long question which is partially answered in many other posts here, but I wanted to bring together everything I know and everything I want to know about the role of the basepoint.

$\endgroup$
1
$\begingroup$

Here are answers to your questions (1)-(3).

  1. What you wrote is correct. The "next best" statement is the following.

Assume that $(X,x_0)$ is well-pointed and let $f,g \colon (X,x_0) \to (Y,y_0)$ be two based maps that are freely homotopic. Then the based homotopy classes $[f]$ and $[g]$ differ by the action of $\pi_1(Y,y_0)$, i.e., there is a loop $\gamma \in \pi_1(Y,y_0)$ with $[g] = \gamma \cdot [f]$. In particular, the induced maps on homotopy groups satisfy $\pi_k(g) = \gamma \cdot \pi_k(f)$ for all $k \geq 1$ (using Hatcher 4.A Exercise 2).

The loop $\gamma$ (or its inverse, depending on your sign convention) is the homotopy class of the path from $f(x_0) = y_0$ to $g(x_0) = y_0$ that traces what the unbased homotopy does to the basepoint.

  1. The following holds irrespective of well-pointedness.

Let $f \colon X \to Y$ be a homotopy equivalence (in the unbased sense). Then $f$ induces a bijection $\pi_0(f) \colon \pi_0(X) \cong \pi_0(Y)$ on path components, and for every basepoint $x_0 \in X$, isomorphisms on homotopy groups $\pi_k(f) \colon \pi_k(X,x_0) \cong \pi_k(Y, f(x_0))$ for $k \geq 1$.

If you started with basepoints $x_0 \in X$ and $y_0 \in Y$ from unrelated path components, then typically $\pi_k(X,x_0)$ would be very different from $\pi_k(Y,y_0)$. But if your setup meant that there is a based map $f \colon (X,x_0) \to (Y,y_0)$ that happens to be a homotopy equivalence (in the unbased sense), then $f$ does induce an isomorphism $\pi_k(f) \colon \pi_k(X,x_0) \cong \pi_k(Y,y_0)$.

  1. Part (2) yields the following.

Let $X$ be contractible (in the unbased sense). Then $X$ is path-connected, and for every basepoint $x_0 \in X$ and $k \geq 1$, we have $\pi_k(X,x_0) = 0$.

Note that this holds even for a degenerate basepoint $x_0 \in X$, e.g., the "origin" in the cone $CH$ on the Hawaiian earring $H$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.