We have two loops $f,g$ in a space $X$ with $f(0)=f(1)=x$ and $g(0)=g(1)=y$ (with $x$ not necessarily equal to $y$), which are homotopic, that is there exists a "path" made of loops (or are any paths allowed?) that transforms $f$ into $g$.
Intuitively I would think that this implies that $x,y$ are path connected, as the loops are continuously deformed into each other, so there must be a "trace" path from $x$ to $y$.
But perhaps the loops are allowed to be "split" among the path components, which would invalidate my intuition.
I ask for clarification which is the case!
I shall expand here my comment: Let $g(t), f(t)$ be two homotopic curves in $X$. Then we have $f([0,1]),g([0,1])$ lie in the same connected component of $X$
Proof: That $f([0,1])$ lie in the same connected component is trivial, the same holds for $g$. Observing that, given the omotopy $F(t,s)$, we have $F(0,0)=f(0);F(0,1)=g(0)$ and $F(0,t)$ is a curve in $X$. Thus $f(0),g(0)$ lie in the same connected component and from transitivity the claim follows