It is easy to see that there is a natural bijection between $\mathcal{C}(X,\Omega Y)$ and $\mathcal{C}(\Sigma X, Y)$, where $\Omega Y$ is the based loop space, $\Sigma X$ is reduced suspension, $X$ and $Y$ are based spaces.
Now $\mathcal{C}(*,*)$ and also $\Omega Y$ can be given the compact-open topology. Is the natural bijection mentioned above actually a homeomorhpism with this topology?
I can prove this is true if I assume $X$ is a compact space. I have shown that there is an obvious map $\Phi : \mathcal{C}(X,\Omega Y)\to \mathcal{C}(X\times S^1, Y)$ and $\Phi$ is a homeomorphism onto its image. From $\text{Im}{\Phi}$ I can define a map into $\mathcal{C}(\Sigma X,Y)$, but to show that this map is continuous I need $X$ to be compact. Then the composition is actually the natural bijection.
Is compactness of $X$ necessary to prove the statement? Can it be generalized to say locally compact spaces or compactly generated spaces?
Any help regarding this is appreciated.
As Rolf Sievers already indicated in the comments, this is true if you work in the category of compactly generated spaces. But be aware that the function spaces do not carry the compact-open topology in this situation, but the compactly generated refinement of it instead.
If you don't want to modify the topology on the function spaces, you have to put stronger conditions on one of the spaces: If $X$ is locally compact, $\Phi$ is a homeomorphism using the regular compact-open topology and any space $Y$. A proof of this appears in Tammo tom Diecks book on algebraic topology (see Theorem 2.4.11 using $\Sigma X\cong S^1\wedge X$).