Help understanding Hatcher's commutative diagram in describing Hopf algebras

I need help understanding the following:

In particular, I need help understanding how $$P$$ is defined. What is $$\mathbb{1}\otimes i^*$$ and then the map pointing upwards into $$H^*(X,R)$$?

I understand that $$P$$ is just the composition of these two maps, but I don't understand what these two maps are. In particular, how does $$P(\alpha\otimes1)=\alpha$$, yet $$P(\alpha\otimes\beta)=0$$ when $$|\beta|>0$$?

This is on page 283 of Hatcher's Algebraic Topology if that helps any.

Thank you.

The map $$1\otimes i^{\ast}$$ is the tensor product of the identity on $$H^{\ast}(X;R)$$ and the map $$H^{\ast}(X;R)\rightarrow H^{\ast}(e;R)$$ induced by the inclusion $$i\colon e\hookrightarrow X$$ (Hatcher uses $$i$$ for two different maps). The right-hand vertical map is the cross product $$H^{\ast}(X;R)\otimes H^{\ast}(e;R)\stackrel{\sim}{\rightarrow}H^{\ast}(X\times e;R)=H^{\ast}(X;R)$$ (the latter identification due to identifying $$X=X\times e$$). Now, following the identification $$X=X\times e$$ by the cartesian product of $$\mathrm{id}_X$$ and the inclusion $$i\colon e\hookrightarrow X$$ yields as composite the inclusion $$i\colon X\rightarrow X\times X$$. Thus, commutativity of the diagram is a consequence of the cross product's naturality. $$P$$ is the composite along this diagram by definition.
Now, $$P(\alpha\otimes1)=\mathrm{id}(\alpha)\times i^{\ast}(1)=\alpha\times1=\alpha$$ since $$i^{\ast}$$ is a ring homomorphism and a standard property of the cross product. On the other hand, $$P(\alpha\otimes\beta)=\mathrm{id}(\alpha)\times i^{\ast}(\beta)=\alpha\times0=0$$ since $$H^n(e;R)=0$$ for $$n>0$$ and the cross product is bilinear.