# Center of fundamental group

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?

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

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.

• Very nice explanation. Commented Apr 30, 2019 at 0:32
• @Coward Did you make any effort to read my answer ?.. Commented Apr 30, 2019 at 16:19

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
$$[\gamma]*[\alpha]=[\alpha]*[\gamma]$$

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 $$\pi_1(X)$$.

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