Karp reduction from PARTITION to SUBSET SUM PARTITION: Given a set of positive integers $A=\{a_1,...,a_n\}$ does there exist a subset of $A$ with sum equal to the sum of it's complement?

SUBSET SUM: Given a set of positive integers $A=\{a_1,...,a_n\}$ and another positive integer $B$, does there exist a subset of $A$ such that it's sum is equal to $B$?

I was trying to prove that if PARTITION is NP-complete then SUBSET SUM is also NP-complete, by reducing PART to SSUM.

My solution was: let $A=\{a_1,...,a_n\}$ be a set of positive integers. Then if A when fed into PART gives the solution $I=\{k_1,...,k_m\}$ (where $k_i$ are the indices of the members of the solution subset), then we construct $A'=\{a_1,...a_n,S\}$ where $S$ is the sum of $\{a_{k_1},a_{k_2},...,a_{k_m}\}$. $A'$ is a solution to SSUM.

My problem with this is that this goes only one way, meaning that we can't show that given A' then A is a solution to PART. Is this a problem? and how could i modify the proof to cover it?
Let's say we have a canditate set $A=\{a_1,...,a_n\}$ to feed as input to PARTITION. There is a transformation in polynomial time $f$, with $f(A)=(A,B)$, where $B=$${\sum_{i=1}^n a_i }\over 2$.
Then $A$ is a solution to $PART$ if and only if $(A,B)$ is a solution to $SSUM$.
So we have $PART \le_m^p SSUM$