# Mayer-Vietoris where $A\cap B$ bounds $A$ and $B$

So I'm a bit confused about how the Mayer-Vietoris Sequence works. I thought that one of the times when it is useful is when we choose $A$ and $B$ such that $A\cap B$ is homotopic to the boundaries of $A$ and $B$, and hence the map $H_n(A\cap B)\to H_n(A)\oplus H_n(B)$ is the zero map since it is induced by the inclusions in each coordinate. I believe this is how MVS is so powerful with computing homologies of connect sums. Let $A$ be a neighborhood around $X$, $B$ a nbd around $Y$, then $A\cap B$ is a circle which is homotopic to the boundaries of $A$ and $B$, and so the map $H_1(A\cap B)\to H_1(A)\oplus H_1(B)$ is the 0 map.

The first issue I have, that I may have resolved but please correct me if I'm wrong, is that it is not the zero map in every dimension, just the one that's $\dim A-1$ hence the dimension of the boundary.

The second issue I'm going to bring up via example. Let's consider the homology of a knot complement. Since it doesn't matter the knot, let's just choose the unknot. Let $A$ be a solid torus containing the knot, let $B$ be the knot complement, hence $A\cap B$ is a torus. The MVS sequence is then $$\dots\to 0\to H_3(S^3)\to H_2(S^1\times S^1)\to H_2(A)\oplus H_2(B)\to H_2(S^3)\to \dots$$ $$\dots\to 0\to \mathbb{Z}\to \mathbb{Z}\to 0\oplus H_2(B)\to 0\to \dots$$

I THOUGHT the way to compute the homology here is to use the fact that the torus bounds both $A$ and $B$ and so the map $H_2(S^1\times S^1)\to H_2(A)\oplus H_2(B)$ was the zero map, and hence $H_2(B)=0$.

The problem I'm having with this is let's instead consider $\mathbb{R}^3-K$ where $K$ is again the unknot and use the same idea for choosing $A$ and $B$. $$\dots\to 0\to H_3(\mathbb{R}^3)\to H_2(S^1\times S^1)\to H_2(A)\oplus H_2(B)\to H_2(\mathbb{R}^3)\to \dots$$ $$\dots\to 0\to 0\to \mathbb{Z}\to 0\oplus H_2(B)\to 0\to \dots$$ and so the map $H_2(S^1\times S^1)\to H_2(A)\oplus H_2(B)$ cannot possibly be the zero map, however the intersection still seems to bound $A$ and $B$.

Something is wrong with my thought process here. Can someone please clarify when we get the map induced by the inclusion is 0 and when it's not?

## 1 Answer

You seem to have a misconception about when the map induced by the inclusions of the intersections is the zero map mainly in regard to what constitutes a boundary. Let's recall how this map is defined. Suppose $i_A\colon A\cap B\to A$ is the inclusion of the intersection into $A$ and $i_B\colon A\cap B\to B$ is the inclusion of the intersection into $B$. The map appearing in the Mayer-Vietoris sequence is defined to be $$((i_A)_*,(i_B)_*)\colon H_n(A\cap B)\to H_n(A)\oplus H_n(B)$$ where $(i_A)_*$ is the induced map on homology of the inclusion map. When is this map the zero map? Well, if it is non-zero on either factor that it will not be the zero map into the sum so each factor of the map must be zero. Hence we need both $(i_A)_*$ and $(i_B)_*$ to be zero. This happens precisely when every $n$-homology class appearing in $A\cap B$ is mapped to an $n$-boundary in $A$ and also in $B$.

We can reduce the work needed by noting that we only need to check cycles which represent the generating elements of the $n$th homology group of $A\cap B$. In your case, looking at $H_2(A\cap B)$ we can consider a single $2$-cycle which represents a generator $2$-homology class of the torus - such a cycle $\alpha$ is given by (some sum of $n$-cells which covers) the torus itself. Now, you know that the inclusion of the torus into $A$ is trivial with respect to the second homology group because $H_2(A)$ is trivial.

So now we need to check with the inclusion into $B$. When $B=S^3\setminus K$ you were right in saying that the $A\cap B$ 'bounds' $B$ but I think you really need to be more careful here. There exists a $3$-chain $\beta$ in $B$ which has the property that $\partial\beta=\alpha$, and this $3$-chain $\beta$ is given by taking exactly one copy of each $3$-cell in the cell complex structure we place on $B$. Note that $\beta$ is a finite sum of $3$-cells and this really comes from the fact that $S^3$ is compact.

If instead $B=\mathbb{R}^3\setminus K$ then we hit a problem. There is no $3$-chain which covers all of $B$ because a $3$-chain needs to be a finite sum of cells, but $\mathbb{R}^3\setminus K$ requires an infinite number of compact cells to cover it. You might think we can get around this by 'shrinking' $B$ down so that $K$ is contained in some closed disk $D$ - after all $B'=D\setminus K$ is a deformation retraction of $B$ so they will have the same homology groups. In this case it's true that we can find a $3$-chain which covers all of $B'$, but now we see that actually $\alpha$ is not the boundary of that $3$-chain because the boundary sphere of the disk $D$ would appear as a summand of such a chain.

In fact we can see from the work that you've already done that $H_2(B)$ will be isomorphic to $\mathbb{Z}$ and indeed a generator of $H_2(B)$ is represented by the image of the generator in $H_2(A\cap B)$. I think the moral of the story here is that 'boundary' in the geometric sense is not necessarily the same as the homological sense, which is really a formal construction which is only somewhat based on our intuition for what a boundary should represent. (However, if everything we're working with is a compact manifold, such as in the case of $S^3\setminus K$, then these two notations of boundary do coincide).