Is the path space of a space homeomorphic to the disjoint union of the path spaces of the path components Let $X$ be an arbitrary wild topological space. Equip the space $\mathcal{C}([0,1],X)$ of continuous paths $[0,1]\rightarrow X$ with the compact open topology. As every path lands in exactly one path component, $\mathcal{C}([0,1],X)$ is in bijection to $\coprod_{X_i\in\pi_0(X)}\mathcal{C}([0,1],X_i)$.
However, is this bijection a homeomorphism?
As $\mathcal{C}([0,1],\_)$ is a functor using the compact open topology, the map from $\coprod_{X_i\in\pi_0(X)}\mathcal{C}([0,1],X_i)$ to $\mathcal{C}([0,1],X)$ is continuous, but is its inverse continuous as well? I assume this surely holds if the path components are open, i.e. $X$ is locally path connected, but does this hold in any case?
 A: No, it is not true if $X$ is not locally path-connected. Consider $X = \{ 1/n \mid n \in \mathbb{N}^* \} \cup \{ 0 \}$ (with the subspace topology $X \subset \mathbb{R}$). Then the path components of $X$ are its singletons, and for each singleton $\{x\}$, $\mathcal{C}([0,1], \{x\})$ is a singleton itself. However, $\mathcal{C}([0,1], X)$ itself is not a countable disjoint union of singleton: it is in fact homeomorphic to $X$ itself, via $$\mathcal{C}([0,1], X) \xrightarrow{\cong} X, \quad f \mapsto f(0)$$
(there is nothing special about $0$: every $f : [0,1] \to X$ is constant, I could have chosen evaluation at any $t \in [0,1]$).
(This can be seen intuitively: if $X$ itself is not homeomorphic to the disjoint union of its path components, then $\mathcal{C}([0,1], X)$ has little chance of being a disjoint union on the path components of $X$ itself. See here for more examples.)

However if $X$ is locally path connected this is true. The map
$$w : \bigsqcup_{X_i \in \pi_0(X)} \mathcal{C}([0,1], X_i) \to \mathcal{C}([0,1], X)$$
is continuous by definition (the inclusion $\mathcal{C}([0,1], X_i) \to \mathcal{C}([0,1], X)$ is continuous, use the universal property of the disjoint union). But it is also open: let
$$C(U,K) = \{ f : [0,1] \to X_i \mid f(K) \subset U \}$$
be an open set in the subbasis defining the topology on $\mathcal{C}([0,1], X_i)$ (i.e. $K \subset [0,1]$ is compact and $U \subset X_i$ is open). Then since $X$ is locally path-connected, its path component $X_i$ is open, and thus $U \subset X_i \subset X$ is open in $X$ too. It follows that $w(C(U,K))$ is also open in $\mathcal{C}([0,1], X)$ (by definition). So $w$ sends open sets of a subbasis to open sets, so it's an open map. To conclude, $w$ is open, continuous, and bijective, so it's a homeomorphism.
