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?
Best Answer
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.