Define $\phi:G/H \rightarrow G/K$ by $\phi(Ha)=Ka$. Prove: $\phi$ is a well defined function. 
Let $H$ and $K$ be normal subgroups of a group $G$, with $H \subseteq K$. Define $\phi:G/H \rightarrow G/K$ by $\phi(Ha)=Ka$.
Prove: $\phi$ is a well defined function. [That is, if $Ha=Hb$, then $\phi(Ha)=\phi(Hb)$.]

Question: To show a function is well defined, do we show that if $x=y$ then $f(x)=f(y)$?  Isn't this the same as an injective function?
Since $\phi(Ha)=Ka$, if $Ha=Hb$, then $Ka=Kb$.  Is this what well defined is? I am not sure what this says.
 A: You have what you need to show right, but that's not the same as an injective function. 
With an injective function, the implication runs if $f(x)=f(y)$ then it must be that $x=y$. 
What we're trying to show here is that if $x$ and $y$ are, in fact, the same thing, then $f(x)$ and f(y) produce the same result regardless of whether we choose to call it $x$ or $y$. Basically, your question asks you to deal with the following worry:
The (possible) function $\phi$ is supposed to map cosets of $H$ it cosets of $K$. However, we can refer to the same coset many different ways: $Ha$ could just as well be written $Hg$ for any other $g\in Ha$. It's not at all obvious, then, that our rule of "map $Ha$ to $Ka$" gives clear instructions. It could be that two perfectly equal ways of writing one coset, $Hg_1=Hg_2$ wind up producing distinct cosets of $K$, $Kg_1\neq Kg_2$. (well, no it couldn't, but that's what you're supposed to prove.) 
Edit: Rather than just describe the worry, it might be easier to show it. $D_4$ has a few normal subgroups; two of them are $A=\{e,r,r^2,r^3\}$ and $B=\{e,s,r^2,sr^2\}$. Let's try to define the homomorphism $\phi$ described in your problem from $D_4/A$ to $D_4/B$. The elements of $D_4/A$ are the cosets $sA$ and $A$. The latter is exactly the same coset as $rA$. Our description of $\phi$ tells us that $\phi(rA)=rB$ or, likewise, $\phi(A)=B$. But the cosets $rB$ and $B$ are not the same, so now we have $rB=\phi(rA)=\phi(A)=B$, but $rB\neq B$, so $\phi$ provides contradictory values for the same input.   
