$x_1$ and $x_2$ lie in the same path component iff $x_1 - x_2 \in im(\Theta)$

Question

This is a question from Rotman's algebraic topology.

Let X be a topological space and let $\Sigma$={ all paths in X }, and let $F(X)$ be the free abelian group on X and $F(X,\Sigma)$ the free abelian group on $\Sigma$. Let $\Theta\colon \Sigma \to F(X)$ be the map which sends a path $\sigma$ to $\sigma(1) -\sigma(0)$. Then we get a unique homomorphism $\Theta\colon F(X,\Sigma) \to F(X)$.

I need to show that

$ x_1 - x_0 \in im \Theta \implies x_1$ and $x_0 $ lie in the same component.

Any hints?

Answer 1

I might be wrong, but it seems this question might not be too bad?

If $x_1 - x_0 \in im \Theta$, then there is a path $\sigma \in \Sigma$ such that $\Theta (\sigma) = x_1 - x_0$. So now we have a path $\sigma$ between the points $x_0$ and $x_1$, i.e. they are path connected, thus must lie in the same path component.