Relative cohomology of open Mobius strip

Question

Let $M=[0,1]\times (0,1)/\sim$ be the open Mobius strip, and consider the compact subspace $A=[0,1]\times[\frac{1}{3}, \frac{2}{3}]/\sim$. I am trying to compute the relative singular cohomology $H^{\ast}(M,M\setminus A)$ using the cohomological long exact sequence of a pair. Denote by $\iota:M\setminus A\rightarrow M$ the inclusion and by $\pi:(M,\varnothing)\rightarrow (M,M\setminus A)$ the map of pairs.

The space $M\setminus A$ is homotopy equivalent to a half open Mobius strip, yes? It therefore is homotopy equivalent to a circle and we find $H^0(M\setminus A)\cong H^1(M\setminus A)\cong \mathbb{Z}$ and $H^i(M\setminus A)=0$ for $i>0.$ For the same reason, $M$ has the same cohomology. But how do I find the induced maps $\iota^{\ast}$ and $\pi^{\ast}$? In homology a continuous maps between path-connected space induced the identity in degree zero, and the other degrees can often be calculated by looking what happens to chains (for example paths if we look at degree one)? Is there a similar way to read off the induced maps (in this case the multiplication constant) in conomology?

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In this case the cohomology groups are just the homology groups by the UCT, but this does nit tell us something about the induced maps, does it?

Answer 1

I assume by "open Mobius strip" you mean the space which is the usual Mobius strip minus its boundary circle. This agrees with $[0,1]\times(0,1)/\sim$ for a suitable sense of $\sim$ - but, in this case, $M\setminus A$ is homotopy equivalent to just a circle and is homeomorphic to an open cylinder; I'm not sure what's meant by a half-open Mobius strip. If you use the reduced LES you'd then find: $$0\to H^1(M;M\setminus A)\to\Bbb Z\to\Bbb Z\to H^2(M;M\setminus A)\to0\to0\to\cdots$$So it all hinges on what the map $\Bbb Z\to\Bbb Z$ is. The UCT tells us that, since the ext term vanishes, the canonical maps $h:H^1(X)\to\hom(H_1(X),\Bbb Z)$ are isomorphisms for $X=M;X=M\setminus A$ (and indeed for any path connected space). Both $M$ and $M\setminus A$ s hare a common loop which generates their first homology; the induced restriction map $H_1(M\setminus A)\to H_1(M)$ is therefore an isomorphism actually it's obvious if you're not being careless that the generating loop of $M\setminus A$ is not a generator in $M$, rather it corresponds to $2$ - and we see that, using the natural isomorphism $h$, the map $\Bbb Z\to\Bbb Z$ is also an isomorphism multiplication by $2$.

It follows $H^1(M;M\setminus A)$ is zero (kernel of $2$) but $H^2(M;M\setminus A)\cong\Bbb Z/2\Bbb Z$.

Alternatively, we can compute the (co)homology of $X:=M/(M\setminus A)$. Noting $A$ itself is a Mobius strip, we observe that $X$ is the result of taking a cylinder $S^1\times I$, collapsing $\{1\}\times I$ to a point, then identifying $[(z,0)]\sim[(-z,1)]$. I will work with the homology of $X$ since I currently find that more appealing but we will see the reduced homology of $X$ is $\Bbb Z/2\Bbb Z$ in degree $1$ and zero elsewhere; the UCT concludes the claim for cohomology.

This can all be made precise by defining suitable quotient maps and using suitable coordinates, but the proof is still clear and certainly more intuitive and less verbose if I may be allowed to appeal to a diagram. In this image, $X$ is as shown; we identify the boundary circles by the different orientations. I apply the Mayer-Vietoris sequence onto the subspaces $A$ and $B$ as shown, where they have been chosen so their borders overlap slightly and $A\cap B\simeq S^1\vee S^1$.

Both $A$ and $B$ deformation retract onto a circle. The reduced Mayer-Vietoris sequence shows $\widetilde{H_\ast}(X)$ is trivial in all degrees except possibly $1$ and $2$. The "interesting fragment" is: $$0\to H_2(X)\to\Bbb Z^2\to\Bbb Z^2\to H_1(X)\to 0$$And as before, everything hinges on the map $\Bbb Z^2\to\Bbb Z^2$. This can be made rigorous but to avoid being really verbose, we can pick the generating loops in $A\cap B\simeq S^1\vee S^1$ to have the same orientation inside $A$; then the first projection of this map is $(x,y)\mapsto x+y$. What are their images in $B$? They are the same up to a precomposition with the reflection map, which is known to have homological degree $(-1)$ for $S^1$ and so it follows the second projection is $(x,y)\mapsto x-y$ (up to a sign depending on the choice of generator of $H_1(B)$).

The kernel of $(x,y)\mapsto(x+y,x-y)$ is easily trivial, so $H_2(X)\cong0$. The image of $(x,y)\mapsto(x+y,x-y)$ consists exactly of the integers $(a,b)$ such that $a+b$ is even, so the cokernel is $H_1(X)\cong\Bbb Z/2\Bbb Z$ as desired.