A Question about Computable Functions Barry Copper states following in his Computability theory book which I have a question about them. 


Exe.4.5.1:   Show that if $\varphi_e(x)  \downarrow $ is a computable relation, then so is  $\varphi_e(x)=y $.
Cor 4.5.3:
    (a) $\varphi_{e,s}(x)  \downarrow $ and $\varphi_{e,s}(x)=y $ are computable relations (of $e,s,x$).
    (b) $\varphi_e(x)  \downarrow  \Leftrightarrow (\exists s) \varphi_{e,s}(x)  \downarrow  $ and $\varphi_e(x)=y \Leftrightarrow (\exists s) \varphi_{e,s}(x)=y$  


Then he states after, "We will see later that $\varphi_e(x)  \downarrow$  is not a computable relation".
My first question is if the latter statement is true then what is the meaning of asking Exe.4.5.1 which would be true trivially.  
Second, we see later that $ K \leq_mK_0=\big\{<x,e>:x\in W_e\big\} $ which we can deduce that $K_0$ is not computable (as $K$ is not computable).  I think also I'm right when I say: $\varphi_{e,s}(x)  \downarrow  \Leftrightarrow (\exists s) \varphi_{e,s}(x)  \downarrow $ and from Cor.4.5.3 (b) we have  $$\varphi_{e}(x)  \downarrow  \Leftrightarrow  \varphi_{e,s}(x)\downarrow .$$
Then using Cor.4.5.3 (a) we can see that -if I'm right- $\varphi_{e}(x)  \downarrow$ is a computable relation too, which leads to a contradiction.  
So, please help me with my questions as I'm in a confliction understanding computable functions.
 A: One of the fundamental concepts in computability theory is the notion of a computable reduction: We have a problem we know is unsolvable (say, the halting problem) and we can use that to show that other problems are unsolvable, by giving a reduction of the halting problem to our new problem. Then, if our new problem was solvable, we can take an instance of the halting problem, computably reduce it to our new problem, solve this, and so solve the halting problem. A contradiction.
So, instead of thinking of the exercise as "$K$ is solvable implies $K_0$ is solvable" is an exercise in noting antecedent is false, try to think you didn't know whether $K$ was solvable or not. You now have a new path to showing it is not solvable: By showing $K_0$ is not a computable set.
More broadly, all the time in my work, I give a reduction from a non-computable set $A$ to another non-computable set $B$. The correct way to intepret this is via Turing machines with oracles: Namely, I have shown that a Turing machine with oracle for $B$ can determine membership in $A$.
As for your question 2 you are not correct in saying that $\varphi_{e,s}(x)\downarrow$ iff $(\exists s)\varphi_{e,s}(x)\downarrow$. Of course $\Rightarrow$ is true, but your claim on the left treats $s$ as fixed, and it does not follow from the fact that $\varphi_{e,s}(x)$ has not converged yet that it could not converge later. Your claimed equivalence actually says that a Turing machine halts iff it halts at stage $0$ which is obviously not true in general.
