Is a path homotopy equivalence class a path component? Let $X$ be a topological space and let $\Omega=\Omega (X;a,b)$ be a path space of $X$ from $a\in X$ to $b\in X$ with a compact-open topology. 
For any $f\in \Omega$, we can consider a homotopy equivalence class $[f]$.
I want to prove or disprove that these classes are path components of $\Omega$.
I first tried to prove it is path connected based on the following sketch, but I'm not sure it is well proved.
Also I 'm stuck on proving $[f]\cup [g]$ is not path connected if $g\notin [f]$.
(EDITED: I added my proof for the converse, but I'm not sure it is valid)
Any help will be appreciated.

$\forall g,h \in [f]$, there exists a continuous function $H:I^2\rightarrow X$ such that $H(t,0)=g(t), H(t,1)=h(t) \ \forall t \in I$
Then we can define $J:I\rightarrow \Omega \ $ by $J(s)(t)=H(t,s)$ and it suffices to show that $J$ is continuous. 
For an open subset $U$ of $X$ and a compact subset $K$ of $I$, let $L_{K,U}=\{f\in \Omega:f(K)\subseteq U\}$.
$\Rightarrow\forall s\in J^{-1}(L_{K,U}), K\times \{s\} \subseteq H^{-1}(U)^{open}\subseteq I^2$
By the tube lemma, $\exists V^{open}$ such that $K\times V \subseteq H^{-1}(U)^{open}$ and $s\in V$. 
Since $V\subseteq J^{-1}(L_{K,U})$,$\ J$ is continuous.

Conversely, suppose there exists a continuous function $J:I\rightarrow \Omega$ such that $J(0)=f$ and $J(1)=g$. Define $H:I^2\rightarrow X$ by $H(t,s)=J(s)(t)$.
Claim $H$ is continuous on $I^2$. 
For given $U^{open} \subseteq X$, consider $(t,s)\in H^{-1}(U)$.
Put $P=J(s)^{-1}(U)^{open} \subseteq I; Q=J^{-1}(L_{K,U})^{open}\subseteq I$ where $K$ is the closure of $P$, a compact subset of $I$. Then $(t,s)\in P\times Q\subseteq H^{-1}(U)$. Thus $H$ is continuous and the equivalence class should be a path component.  
 A: For any spaces $X$, $Y$, and $Z$, a map $f:X\times Y\to Z$ induces a map $\hat f:X\to Z^Y$, where $Z^Y$ denotes the space of maps $Y\to Z$ equipped with the compact-open topology. The proof you gave for the case $X=Y=I$ applies also to the general case, and the main ingredient is the tube lemma.
Now given a map $g:X\to Z^Y$, we may define the function $\bar g:X\times Y\to Z,\, (x,y)\mapsto g(x)(y)$. This function, however, need not be a continuous map. If we want any map $g$ to imply a map $\bar g$, then setting $X=Z^Y$ and $g=1_X$, we see that $\overline{1_X}:(f,y)\mapsto 1_X(f)(y)=f(y)$ must be continuous. On the other hand, if $\varepsilon:(f,y)\mapsto f(y)$, is a map, then given any map $g$, this induces a continuous $\bar g$, as we may write $\bar g$ as the composite $\varepsilon(g\times 1_Y)$.
The proof you gave in the comments already captured the main reason for $\varepsilon$ to be continuous, namely that when $f(y)\in U$, we can find a compact neighborhood $W$ of $y$ such that $f(W)\subseteq U$, and then $L_{W,U}\times W$ is the desired neighborhood of $(f,y)$ such that $\varepsilon$ sends this to $U$.
