Fixing the representative of a coset for a function. I am working on through a proof from lecture. Given that $K\leq H\leq G$, we are constructing a bijection $$f: G/H\times H/K\rightarrow G/K$$
Since $G/H$ is the set of cosets, we can fix the representative of the cosets in creating a well defined function. That is $$g_1H = g_2H$$
if and only if $g_1 = g_2$. 
That last part about fixing the representatives of each coset leaves me uneasy. Since we know that multiple representatives prevents us from creating an operation on any set of cosets, I am not sure why we can fix our representatives. 
If that is true then the rest of the proof follows.Why are we allowed fix representatives of the cosets for this function. Is it related to mapping cosets from one set to a different set?
 A: As I said in the comments above: $f$ will depend on the choice of the set of representatives.
I think the best way to see what's going on is with an example, so let


*

*$G = \{e, g, g^2, g^3, g^4, g^5\}$, a cyclic group of order 6,

*$H = \{e, g^3\}$, the unique subgroup of order 2,

*$K = \{e\}$.


$G/H$ is a cyclic group of order 3, and we could pick a bunch of different lists of our favourite representatives. Two possible choices are, for example,


*

*$R^{G/H}_1 = \{e, g, g^2\}$

*$R^{G/H}_2 = \{e, g, g^5\}$
-- that is, I've just swapped the $g^2$ for a $g^5$.
$H/K$ is a cyclic group of order 2, and only has one possible set of representatives: $\{e, g^3\}$. (This is because I've taken $K$ to be trivial, for the sake of simplicity, so that we can just ignore it in this example.) So we're forced to set both $R^{H/K}_1$ and $R^{H/K}_2$ -- our lists of our chosen representatives of $H/K$ -- equal to this set.
We can now define our functions
$$f_1, f_2: G/H \times H/K \to G/K$$
as follows: given some cosets $xH\in G/H$ and $yK\in H/K$, the function $f_1$ will look up our chosen representative for $xH$ in the list $R^{G/H}_1$, and our chosen representative for $yK$ in $R^{H/K}_1$; and $f_2$ will do the same, but using the lists $R^*_2$. Both functions then proceed to multiply these elements in order and stick a $K$ at the end and output the result.
So, for instance: consider the cosets $(eH, eK)$. $f_1$ will look up the representatives in lists $R^*_1$, and find that they're $e$ and $e$; $f_2$ will look up the representatives in $R^*_2$ and get the same answers. The result will be $f_1(eH, eK) = f_2(eH, eK) = e\cdot e\cdot K = eK$.
But consider the cosets $(g^2H, eK)$. When $f_1$ looks up the representatives, it finds $g^2$ and $e$; but when $f_2$ looks them up, it finds $g^5$ and $e$. So $f_1(g^2H, eK) = g^2K$, but $f_2(g^2H, eK) = g^5K$. These aren't equal (as elements of $G/K$).
You can visualise the outputs of $f_1$ and $f_2$ by drawing a table for each: 3 rows (one for each coset of $G/H$), two columns (for cosets of $H/K$). You'll notice that, in both cases, all 6 outputs are distinct - in other words, both are injective.
