Show this homotopy is continuous with respect to the compact open topology. Let $\Omega X$ be the function space of $S^1 \to X$, topologize this with the compact open topology (i.e., the open sets are generated by sets consisting elements $f$ such that $f(K)\subset U$ for some $K$ compact in $S^1$ and $U$ open in $X$).
I want to show that the following homotopy is continuous:
$H:\Omega X \times I \to \Omega X$
$ H(g(t),s)=$ $$t\mapsto\left \{
\begin{array}{ll}
      g(2ts) & 0<t<0.5 \\
      g((2-2t)s)& 0.5<t<1 \\
\end{array} 
\right. $$
So, this is the homopoty from composing $g$ with its inverse to the constant map. But I don't see why it is continuous easily.
 A: As a general rule, one should try to avoid using the compact-open topology directly. Instead, use the following two facts:


*

*A function $F: X \to \mathrm{Map}(Y,Z)$ is continuous if and only if the adjoint map $f: X \times Y \to Z$ defined by $f(x,y) = F(x)(y)$ is continuous.

*The evaluation map $Y \times \mathrm{Map}(Y,Z) \to Z$ is continuous.


These are the first two facts you prove about the compact open topology, and it's the whole reason we use it in the first place. Once you've proven these facts once, there's no sense repeating work you've already done, so you try to reduce stuff to this.
So first, to check that $H$ is continuous, need only check that the map $h: \Omega X \times I \times I \to X$ given by $h(g, s, t) = H(g,s)(t)$ is continuous. Now, the definition of $H$ is piecewise, and to check a piecewise defined function is continuous we need only check that each piece is continuous and that it's well-defined on the overlap. Well-definedness is okay, so we need to check that the maps $\Omega X \times I \times [0,1/2] \to X$ and $\Omega X \times I \times [1/2, 1]$ are continuous.
To do that, we'll write each as a composite of a bunch of continuous functions, one of which will be evaluation. The first map can be written as the composite: $\Omega X \times I \times [0,1/2] \to \Omega X \times I \to X$ where the first map is the product of the identity map on $\Omega X$ and the multiplication map $I \times [0,1/2] \to I$ and the second map is evaluation. Since multiplication is continuous, we're good.
The second piece is similar except we use a mixture of multiplication and subtraction before evaluating, but those are continuous operations, so we're good.
