Show that if $x_1, x_0 \in X$, then $x_1 - x_0 \in \operatorname{im}(\partial_1)$ if and only if $x_0, x_1$ lie in the same path component of $X$. 
Let $X$ be a topological space and $S_1(X)$ the free abelian group of paths $\sigma : I \to X$. Let $S_0(X)$ be the free abelian group of points in $X$. Show that if $x_1, x_0 \in X$, then $x_1 - x_0 \in \operatorname{im}(\partial_1)$ if and only if $x_0, x_1$ lie in the same path component of $X$. Here $\partial_1:S_1(X) \to S_0(X)$ is defined by $\partial_1(\sigma)=\sigma(1)-\sigma(0)$.

For the first direction suppose that $x_1 -x_0 \in \operatorname{im}(\partial_1)$, then $$x_1-x_0=\partial_1(\sigma) = \sigma(1)-\sigma(0)$$ which implies that $\sigma$ is a path connecting $x_0$ and $x_1$.
If $x_0$ and $x_1$ are in the same path component, then $\exists \gamma:I \to X$ such that $\gamma(0) =x_0$ and $\gamma(1)= x_1$  and $\gamma$ is continuous. Thus $x_1-x_0 = \gamma(1)-\gamma(0) = \partial_1(\gamma)$.
I'm not sure that the forward direction is correct. What I wanted to conclude that the points are in the image $\operatorname{im}(\partial_1)$, then there exists a path connecting them which would imply that they are in the same path component, but $$x_1-x_0 = \sigma(1) - \sigma(0) \text{ does not imply that } x_1=\sigma(1) \text{ and } x_0 = \sigma(0).$$
 A: We must remember the Abelian group (I think of this like a vector space) $S_0(X)$ is rather unusual in that it is free. The elements are finite formal sums of some generators $(x)_{x\in X}$, and: $$x_1-x_0=\sum_{k=1}^nc_ky_k$$For some generators $(y_k)\in X$ if and only if there are $k,k'$ with $c_k=1,y_k=x_1,c_{k'}=-1,y_{k'}=x_0$ and all other elements are zero. There are essentially no relations in a free Abelian group; there is no chance that $y+\text{(something not including $-y$)}=x$ for $y\neq x$.
We know $\sigma(1)-\sigma(0)=\partial_1(\sigma)=x_1-x_0$ for some $\sigma\in S_1(X)$.
But by the above remarks, the two are equal only if the generators $x_1,x_0$ are present and all other generators have a coefficient of zero. Either $\sigma(1)=x_1,\sigma(0)=x_0$ or $\sigma(1)=-x_1,\sigma(0)=-x_0$. By potentially reversing time in $\sigma$, $\sigma$ must provide a connection $x_0\to x_1$.
A: Unless $x_1=x_0$ (in which case there is nothing to prove) it does imply that $x_1=\sigma(1)$ and $x_0 = \sigma(0)$ because $S_0(X)$ is the free abelian group whose generators are the points of $X$. On both sides of the equation we have two distinct generators, one with a coefficient $+1$, the other with a coefficient $-1$. By the uniqueness of representations of elements of $S_0(X)$ as linear combinations of the basis elements we are done.
A: You are nearly right in your doubts. More precisely, $x_1-x_0\in\operatorname{im}(\partial_1)$ does not mean a priori that $x_0=\sigma(0)$ and $x_1=\sigma(1)$ for a single path $\sigma:I\to X,$ but only that
$$x_1-x_0=\sum_{\sigma\in S}n_\sigma\left(\sigma(1)-\sigma(0)\right)$$
for some finite set $S$ of paths and some family $(n_\sigma)_{\sigma\in S}$ of relative integers.
The standard method for proving that $x_0$ and $x_1$ are in fact in the same path-connected component is given in The zeroth homology group corresponds to path components. Two more "pedestrian" proofs (one handwaving, another one more formal) were given in Difference $x-y$ lies in $\operatorname{Im}(\partial_1)$ if and only if $x,y$ lie in the same path component.
