I need help understanding the following:

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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.


1 Answer 1


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.


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