I am looking at lemma 3.36 in the book Algebraic Topology by Hatcher (p. 246). In the proof we need to show commutativity of a certain diagram and Hatcher assumes two squares to commute fairly trivially. I tried to work the first one out but ran into some trouble.

Assume $M$ is an $R$-oriented manifold that is the union of two open sets $U$ and $V$. $K\subset U, L\subset V$ are compact sets.

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The horizontal arrows come from Mayer-Vietoris sequences and they are just the inclusion induced morphism to each component (with a minus sign in front of the second one). The two isomorphisms come from excision and are also inclusion induced morphisms. The bottom two vertical arrows come from the cap product with $\mu_{K\cap L}\in H_n(U\cap V,(U\cap V)\setminus (K\cap L)),\mu_K\in H_n(U,U\setminus K),\mu_L\in H_n(V,V\setminus L)$, respectively. We define for example $\mu_K$ as the unique element in $H_n(U,U\setminus K)$ for $x\in K$ gets sent to the orientation on $U$ induced from $M$ by the inclusion induced morphism $$H_n(U,U\setminus K)\to H_n(U,U\setminus \{x\})$$ $\mu_{K\cap L}$ and $\mu_L$ are defined in the same way.

Now the reason why I am having trouble is because the map from $H^k(U\cap V,(U\cap V)\setminus (K\cap L))$ to $H^k(U,U\setminus K)$ (or the other component) is not induced from an inclusion (or any natural map afaik). Therefore I am unable to really use the naturality property of the cap product to show this diagram commutes. Does anyone have any insight for me?


1 Answer 1


$\textbf{Lemma:}$ Let $g\colon (X,A,B)\to (\overline X,\overline A,\overline B)$ be a map between two excisive triads, then for $x\in H^k(\overline X,\overline A), y\in H_l(X,A\cup B)$ we have $$x\frown g_*(y)=g_*\big(g^*(x)\frown y\big)\in H_{l-k}(\overline X,\overline B).$$

enter image description here

Commutativity of the red portion: The inclusion map $i\colon(U,U\backslash K)\hookrightarrow (M,M\backslash K\cap L)$ is the composition of two inclusion maps $$(U,U\backslash K)\overset{\beta}{\hookrightarrow}(U,U\backslash K\cap L)\overset{f}{\hookrightarrow}(M,M\backslash K\cap L)$$ Similarly, the inclusion map $i\colon (U\cap V,U\cap V\backslash K\cap L)\hookrightarrow(M,M\backslash K\cap L)$ is the composition of two inclusion maps $$(U\cap V,U\cap V\backslash K\cap L)\overset{\alpha}{\hookrightarrow}(U,U\backslash K\cap L)\overset{f}{\hookrightarrow}(M,M\backslash K\cap L)$$ We first show that $$H^k(M,M\backslash K\cap L)\xrightarrow{(f\circ \beta)^*}H^k(U,U\backslash K)\xrightarrow{\bullet\frown \mu_K^U} H_{n-k}(U)$$ is same as $$H^k(M,M\backslash K\cap L)\xrightarrow{(f\circ \alpha)^*}H^k(U\cap V,U\cap V\backslash K\cap L)\xrightarrow{\bullet\frown \mu_{K\cap L}^{U\cap V}}H_{n-k}(U\cap V)\xrightarrow{\alpha_*}H_{n-k}(U)$$ are the same.

So take $x\in H^k(M,M\backslash K\cap L)$ and consider $$\alpha_*\bigg((f\circ \alpha)^*(x)\frown \mu_{K\cap L}^{U\cap V}\bigg)=\alpha_*\bigg(\alpha^*\big(f^*(x)\big)\frown \mu_{K\cap L}^{U\cap V}\bigg)$$$$=f^*(x)\frown\alpha_*\left(\mu_{K\cap L}^{U\cap V}\right)=f^*(x)\frown \mu^U_{K\cap L}$$ Note that in the last equality, we are using the following uniqueness property:

For an $n$-dimensional oriented manifold $N$ and $A\subseteq_\text{compact}N$ and for every section $\{\varphi_x\}_{x\in A}$ we have a unique $\alpha_A\in H_n(N,N\backslash A)$ such that inclusion induced map $H_n(N,N\backslash A)\to H_n(N,N\backslash x)$ sends $\alpha_A$ to $\alpha_x$ for each $x\in A$.

A similar logic gives $$\beta_*\bigg((f\circ \beta)^*(x)\frown \mu_{K}^{U}\bigg)=\beta_*\bigg(\beta^*\big(f^*(x)\big)\frown \mu_{K}^{U}\bigg)=f^*(x)\frown\beta_*\left(\mu_{K}^{U}\right)$$

Now, $\beta\colon U\hookrightarrow U$ is identity. So, $\text{Id}=\beta_*\colon H_{n-k}(U)\to H_{n-k}(U)$ hence comparing the left most and the right most sides of the above equality we have $$(f\circ \beta)^*(x)\frown \mu_{K}^{U}=f^*(x)\frown\beta_*\left(\mu_{K}^{U}\right).$$ Again, the above uniqueness property shows $$\beta_*\left(\mu_K^U\right)=\mu_{K\cap L}^U.$$ Therefore, combining all these

$$\alpha_*\bigg((f\circ \alpha)^*(x)\frown \mu_{K\cap L}^{U\cap V}\bigg)=f^*(x)\frown \mu^U_{K\cap L}$$$$=f^*(x)\frown\beta_*\left(\mu_{K}^{U}\right)$$$$=(f\circ \beta)^*(x)\frown \mu_{K}^{U}$$ We have shown the commutativity of a portion(corresponding to left summand, red shaded) of the pentagon of the above diagram.

Similarly, one can show the commutativity of the other portion(corresponding to the right summand) of the pentagon of the above diagram. The only thing left is the commutativity topmost triangle, which is easy as all maps are inclusion induced.

  • $\begingroup$ Thanks for the detailed answer! One thing I am still wondering is: we have commutativity for the right summand because the bottom map is -i and on the right we take the negative cap product so these signs cancel. However, showing that the square to the right of this one commutes, we also take the negative cap product on the left, but I do not see any signs showing up on the right. Any idea where I went wrong? $\endgroup$ May 25, 2021 at 11:22
  • $\begingroup$ Mine is $$H^k(M,M\backslash K\cap L)\xrightarrow{-i}H^k(V,V\backslash L)\xrightarrow{\bullet\frown \mu_K^U} H_{n-k}(V)$$ is same as $$H^k(M,M\backslash K\cap L)\xrightarrow{\cong}H^k(U\cap V,U\cap V\backslash K\cap L)\xrightarrow{\bullet\frown \mu_{K\cap L}^{U\cap V}}H_{n-k}(U\cap V)\xrightarrow{-i}H_{n-k}(V)$$ are the same. $\endgroup$
    – Sumanta
    May 25, 2021 at 12:02
  • $\begingroup$ Hatcher's is $$H^k(M,M\backslash K\cap L)\xrightarrow{i}H^k(V,V\backslash L)\xrightarrow{-\bullet\frown \mu_K^U} H_{n-k}(V)$$ is same as $$H^k(M,M\backslash K\cap L)\xrightarrow{\cong}H^k(U\cap V,U\cap V\backslash K\cap L)\xrightarrow{-\bullet\frown \mu_{K\cap L}^{U\cap V}}H_{n-k}(U\cap V)\xrightarrow{i}H_{n-k}(V)$$ are the same. $\endgroup$
    – Sumanta
    May 25, 2021 at 12:04
  • 1
    $\begingroup$ You can consider either of these two but do not combine them. Here, $-\bullet\frown \mu_K^U$ is the inverse of the homomorphism $\bullet\frown \mu_K^U$, and similarly for the other. $\endgroup$
    – Sumanta
    May 25, 2021 at 12:12
  • $\begingroup$ Ok, thank you very much! I guess I was misinterpreting the maps from Hatcher's diagram $\endgroup$ May 25, 2021 at 12:51

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