3
$\begingroup$

Let $f_t: X \rightarrow X$ be a homotopy of maps such that $f_0 = f_1 = \mathrm{id}_X$. For any $x_0 \in X$, the map $t \mapsto f_t(x_0)$ is a loop based at $x_0$.

To prove: $[f_t(x_0)]$ is contained in the center of $\pi(X, x_0)$.

Attempt:

If $y$ is an element of $\pi(X, x_0)$, then $y$ commutes with $[f_t(x_0)]$ because $[f_t(x_0)]$ is homotopic to the identity. That is, we can replace $[f_t(x_0)]$ with the identity loop, since they are in the same class, and the identity loop is in the center of $\pi(X, x_0)$.

Thoughts?

$\endgroup$
3
  • $\begingroup$ What if $X=S^1$ and $f_t$ is rotation by $2\pi t$. Why is $[f_t(x_0)]$ homotopic to the identity? $\endgroup$ Commented Apr 28, 2019 at 23:42
  • $\begingroup$ @MichaelBurr Informally, because you can rotate by $2\pi t$ in the other direction, which ends at the identity. $\endgroup$
    – Coward
    Commented Apr 28, 2019 at 23:46
  • 3
    $\begingroup$ What Michael is saying is $F_t(x)=e^{ix} \mapsto e^{i(x+t)}$ then $F_t(x_0)$ isn't trivial in $\pi_1(S^1)$ $\endgroup$
    – reuns
    Commented Apr 29, 2019 at 1:11

3 Answers 3

6
$\begingroup$

Let $p:[0,1] \to X$ be a loop based at $x_0$.

The map $h(s,t) = f_t(p(s))$ is a continuous mapping $[0,1] \times [0,1] \to X$.

On the boundary of the square, the following is true:

$h(0,t) = f_t(x_0) = h(1,t) \quad$ and $\quad h(s,0) = p(s) = h(s,1)$.

In other words, $h$ restricts to $p(s)$ along the bottom and top sides and restricts to $f_t$ along the left and right sides. The concatenation of two paths, $f_t*p(s)$, is the left side followed by the top side. The concatenation of two paths, $p(s)*f_t$, is the bottom side followed by the right side. That $h$ extends this pair of concatenated paths over the square shows that they are homotopic relative to their endpoints.

If you sketch this square and then draw a diagonal from the upper left to the lower left, then you can visualize the homotopy (rel endpoints) by restricting $h$ to the family of line segments which join the lower left vertex to a point on the diagonal and then to the upper right vertex.

homotopy square

$\endgroup$
2
  • 1
    $\begingroup$ Very nice explanation. $\endgroup$
    – Coward
    Commented Apr 30, 2019 at 0:32
  • $\begingroup$ @Coward Did you make any effort to read my answer ?.. $\endgroup$
    – reuns
    Commented Apr 30, 2019 at 16:19
0
$\begingroup$

I'd like to give a proof in which we don't explicitly construct the homotopy we need to conclude, using the fact that the square is contractible, which is a more abstract property (and easier to explicitly show) than having the sides homotopic to the diagonal of the square.

In the following, we denote $[0,1]$ by $I$
Let $F:X\times I \rightarrow X$ be our initial homotopy and $\gamma = F(x_0,\cdot)$

For any loop $\alpha$ based in $x_0$, we can define the function $H=F(\alpha(\cdot),\cdot):I\times I\rightarrow X$ (which is continuous) And we can easily see that $H(\cdot,0)=H(\cdot,1)=Id_X\circ\alpha=\alpha$ and $H(0,\cdot)=H(1,\cdot)=F(x_0,\cdot)=\gamma$

Let $H_*:\Pi_1(I\times I)\rightarrow\Pi_1(X)$ the functor between fundamental groupoids induced by $H$
Let $a,b,c,d$ be the paths along each side of the square (turning clockwise)

Again, we can easily see that:
$H\circ a=\gamma\quad H\circ c=\gamma^{-1}\quad H\circ b=\alpha\quad H\circ d=\alpha^{-1}$
Or equivalently,
$H_*[a]=[\gamma]\quad H_*[c]=[\gamma]^{-1}\quad H_*[b]=[\alpha]\quad H_*[d]=[\alpha]^{-1}$

But we know that $I\times I$ is contractible and $a*b*c*d$ is a loop based at $0$
Hence, we have:
$[a]*[b]*[c]*[d]=[0]$
And so
$H_*[a*b*c*d]=[\gamma]*[\alpha]*[\gamma]^{-1}*[\alpha]^{-1}=[x_0]=H_*[0]$
Which is equivalent to
$[\gamma]*[\alpha]=[\alpha]*[\gamma]$

$\endgroup$
-1
$\begingroup$

Let $C(X)$ be the set of curves in $X$ and $Hom(X)$ the homeomorphisms $X \to X$.

For $F : id_X\to id_X \in C(Hom(X))$ and some $x_0 \in X$ let $f_t = F_t(x_0)$ so that $f \in C(X)$.

For any curve $g:x_0 \to x_0 \in C(X)$ let $h_t = ( f_\tau,\tau \in [0,t]) \cup F_t(g) \cup ( f_{t-\tau},\tau \in [0,t])$ so that $h \in C(C(X))$.

$h_0 = g$ and $h_1 = f \cup g \cup f_{-}$ and $h_t$ is an homotopy between the two.

Thus we obtained that $f$ commutes with $g$ in $\pi_1(X)$.

$\endgroup$
2
  • $\begingroup$ Thanks. Can you use mostly words to describe what you are saying with all the symbols? It went way over my head. $\endgroup$
    – Coward
    Commented Apr 29, 2019 at 1:38
  • $\begingroup$ Look carefully at the definition of $h_t$, the concatenation of 3 curves, interpolating between $g$ and $f \cup g \cup f_-$ $\endgroup$
    – reuns
    Commented Apr 29, 2019 at 1:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .