On proposition I.3.2 of 'Quantum groups' by Kassel. I am reading the book Quantum groups by Kassel. In proposition I.3.2 at the very beginning the reader is asked to show that under the identifications made, the maps $\Delta,\varepsilon$ and $S$ correspond to the maps $+,0$ and $-$.
However, $+$ is a map from $A^2$ to $A$, and I'm not sure how we can use the identifications to see that $\Delta:k[x]\rightarrow k[x',x'']$ corresponds to $+$.
It seems to me, all you can get from the identifications is that the map $\Delta$ corresponds to the element $x'+x''\in k[x',x'']$. Similarly, $\varepsilon$ corresponds to the element $0\in k$ and $S$ corresponds to the element $-x\in k[x]$.
So what exactly is meant in this proposition? I'm fairly sure this is a stupid question, but it's one that should be well-understood before proceeding any further in this theory.
Thank you in advance.

The functor $\mathrm{Hom}(-,A)$ is a contravariant functor, so a map $f:k[x]\to k[x',x'']$ corresponds to a map $\mathrm{Hom}(k[x',x''],A)\to\mathrm{Hom}(k[x],A)$ given by pre-composing with $f$. If you write down what this means for $\Delta$ you start with a map $k[x',x'']\to A$ that takes for instance $x'\mapsto a$ and $x''\mapsto b$, corresponding to $(a,b)\in A^2$. Pre-composing with $\Delta$ we get the map $x\mapsto a+b$, which corresponds to $a+b\in A$. So we see that $\Delta$ corresponds to the map $(a,b)\mapsto a+b$. You can do the same to get the correspondences for $S$ and $\epsilon$.
