Difference $x-y$ lies in $\operatorname{Im}(\partial_1)$ if and only if $x,y$ lie in the same path component Exercise

For a given space $X$, define $S_1(X)$ to be the free abelian group with basis all paths $\sigma \colon \mathbf{I} \to X$, and let $S_0(X)$ be the free abelian group with basix $X$.
If $x_1,x_0\in X$, show that $x_1 - x_0 \in \operatorname{im}\partial_1$ if and only if $x_0,x_1$ lie in the same path component of $X$.

I want to prove the implication to the right of part 2. 
If $x_1-x_0$ is a boundary for $\operatorname{\partial_1}$, I now how to prove that $x_0,x_1$ lie in the same path component heuristicly for cases  being the boundary of a finite linear combination of length 1, 2 or 3. 
That is: assume $x_1-x_0=\partial_1  \ \left(\lambda_1 (\sigma_1(1)-\sigma_1(0))+...+\lambda_3(\sigma_3(1)-\sigma_3(0))\right)$. Then I know how to prove it. But don't see how to prove in general, or how to get to the inductive step. 
 A: (Easy direction)
Suppose $x_0,x_1$ lie in the same path component and let $\sigma$ be a path from $x_0$ to $x_1$. Then $\sigma(0)=x_0$ and $\sigma(1)=x_1$ and so $\partial_1(\sigma)=x_1-x_0$. It follows that $x_1 - x_0 \in\operatorname{im}\partial_1$.
(Hard direction)
Suppose $x_1 - x_0 \in \operatorname{im}\partial_1$. Then there exists a finite set of paths $\sigma_i \colon\mathbf{I}\to X$, $i \in \{1, \ldots, k\}$ and coefficients $a_i$ such that $\partial_1 \sum a_i\sigma_i = x_1 - x_0$.
So, $$\begin{align}\sum a_i\sigma_i(1)-a_i\sigma_i(0) = x_1 - x_0\end{align}\tag{1}$$ by definition of $\partial_1$. Without loss of generality, we may assume that each of the $a_i$ is non-negative, as otherwise, replace $a_i$ with $-a_i$ and $\sigma(t)$ with $\sigma(1-t)$.
Let $A = \sum a_i$. If $A = 1$, then we have $\partial_1 \sigma = \sigma(1) - \sigma(0)$, so $\sigma(1) - \sigma(0) = x_1 - x_0$. By the fact that $S_0(X)$ is a free group generated by elements of $X$, we must have $\sigma(0) = x_0$ and $\sigma(1) = x_1$. So, $\sigma$ is a path from $x_0$ to $x_1$, hence they lie in the same path component. This is out base case.
Now in general, assume that $\sigma_1(0) = x_0$ (otherwise, reorder the $\sigma_i$s). Either $\sigma_1(1) = x_1$, in which case we're done, or else $\sigma_1(1) = v$ for some $v \neq x_1$. It follows that $\sum a_i \sigma_i(1)$ contains a positive $v$ component. So, there must be some $\sigma_j \neq \sigma_1$ such that $\sigma_j(0) = v$, as the right hand side of equation $(1)$ has no $v$ component. Without loss of generality, assume that $j=k$. Then, in the sum, we may replace one copy of $\sigma_0 + \sigma_k$ with their concatenation $\sigma_0\# \sigma_k$, as $\sigma_{k+1} = \sigma_0(1) = \sigma_k(0)$. This means we've replaced the sum $\partial_1 \sum_{i=1}^k a_i\sigma_i = x_1 - x_0$ with the sum
$$\partial_1 \left((a_1-1)\sigma_1 + (a_k-1) \sigma_k + \sigma_{k+1} + \sum_{i=2}^{k-1} a_i\sigma_i\right) = x_1 - x_0.$$
The new sum has $$A' = a_1-1 + a_k-1 +1 + \sum_{i=2}^{k-1} a_i\sigma_i = A-2+1 = A-1.$$
So we've reduced the sum of the coefficients by $1$. We can repeat the process until $A' =1$, in which case we've reached the base case and we're done.

Probably the important part to remember for the left to right implication is that free abelian groups have the nice property that if $n_1g_1 + \cdots n_k g_k = m_1h_1 + \cdots m_lh_l$ for integers $n_i,m_i$ and generators $g_i,h_i$, and the $n_i$ are all pairwise distinct, then $k=l$, the $m_i$ are all pairwise distinct also, and $g_i=h_j$ if and only if $n_i = m_j$.
A: Think about this, for a linear combination of paths, we act boundary operator on it, then we get a collection of vertices containing the $v_1$ and $v_0$, notice in this collection we also count multiplicity i.e. when a vertice belongs to several paths then we count them repetitively, this collection also preserve information of the sign of the vertice(to be an end or to be a start) let's call the elements signed vertice(0-simplex), now we know this collection's signs should be balanced off altogether because every path produces a balanced pair of signs.This result clearly hold for any subset of that collection provided this subset is the signed vertices collection of a path subset of the original path combination which introduces the first collection. now let's think about the "path component" containing $v_1$ and $v_0$ i.e. the collection respecting multiplicity of all the signed vertices able to use the path in the combination to connect to $v_1(v_0)$, if they are separated then they will independently respectively produces a balanced positive and negative signs and most importantly they can not be cancelled out(cancel of 0-simplex!) with the signed vertices in the other components as a special case $v_0$ set for $v_1$, $v_1$ set for $v_0$ because they are separated(of course there maybe are components other than the ones containing $v_1$ and $v_0$ but same idea: the signed vertices of component can only be cancelled in itself's set). so the addtional signed vertices can remain itself's existence when counting it in the overall boundary operation of the whole path combination but the result of the boundary operator acting on the whole 1-chain in $S_1(X)$ gives only $v_1-v_0$, i.e. nothing other than $v_1,v_0$ remain itself. What is that mean?  $v_0(v_1)$'s "path component" will give us a odd number of signs (i.e. a single + and a single -) that can't be balanced off by itself. CONTRADICTION
