I want to prove that if $E'$ is a completion of a normed space $E$ then $E'$ is a Banach space. (with isometry $i$)
In my lecture the professor gave the following proof: First we can extend the $+$, $\cdot$ and $|| \cdot ||$ to $E'$ using that the diagrams commute (I'm only writing the one for the sum because it ilustrates the point of my doubt)
$E \times E \overset{i\times i}{\longrightarrow} i(E)\times i(E)$
$\space \space +$ $\downarrow$ $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space\downarrow +'$
$\space \space \space \space \space E \space \space \space \space \space \overset{i}{\longrightarrow} \space \space \space \space \space \space \space \space E'$
Using that the diagram commutes it's simple to see that there is a unique extension of the $+$, $\cdot$ and $|| \cdot||$ to $+'$, $\cdot'$ and $||\cdot||'$ and then proving that $E'$ is normed is simple.
My doubt is why can we say that the diagram commutes, it would be true if the isometry $i$ is linear but I'm sure why we could say that.
Note that when you start the proof, no operator $+': i(E) \times i(E) \to E'$ is defined.
What is true is that there is a unique (uniformly continuous) operator $+': i(E) \times i(E) \to E'$ such that the diagram you've drawn commutes. I imagine that your professor intended you to take $+'$ to be that operator. Then, since $i(E) \times i(E)$ is dense in $E' \times E'$, we can uniquely extend $+'$ to a continuous operator $E' \times E' \to E'$.