Let $K\_0$ and $ K\_1$ be knots in $S^3$. They are called *smoothly concordant* if there is a smoothly properly embedded cylinder $S^1 \times [0,1]$ in $S^3 \times [0,1]$ such that $\partial (S^1 \times [0,1]) = -(K\_0) \cup K\_1$.


Let $Y\_0$ and $ Y\_1$ be integral homology spheres, i.e., $H\_\*(Y\_i; \mathbb Z) = H\_\*(S^3; \mathbb Z)$. They are called *homology cobordant* if there exists a smooth compact oriented $4$-manifold such that $\partial X = -(Y\_0) \cup Y\_1$ and $H\_\*(X,Y\_i; \mathbb Z)=0$ for $i=0,1$.


I cannot explicitly figure out but I made some progress. How can we concretely prove that the following well-known theorem: Let $S\_n^3(K)$ denotes $3$-manifold obtained by the $n$-surgery on the knot $K$ in $S^3$.


**Theorem:** If $K\_0$ is smoothly concordant to $K\_1$ in $S^3$, then for all $n$, $S\_n^3(K\_0)$ is homology cobordant to $S\_n^3(K\_1)$.


**Addition:** Can we use this theorem to obtain "strong" obstructions for knots being smoothly concordant?



Relating smooth concordance and homology cobordism via integral surgeries
I will call $X\_n(K)$ the trace of $n$-surgery along $K$, that is a 4-manifold diffeomorphic to the union of $B^4$ and an $n$-framed 2-handle attached along $K \subset S^3 = \partial B^4$.


Call $A \subset S^3 \times I$ the concordance from $K\_0$ to $K\_1$.
Consider $X\_1 := X\_n(K\_1)$, viewed as $B^4 \cup S^3\times I \cup H$, where $H$ is the 2-handle. For convenience, I will call $C$ the core of $H$. I claim that $X\_n(K\_0)$ embeds in $X\_n(K\_1)$ as a regular neighbourhood, that I'll call $X\_0$, of $B^4 \cup A \cup C$. This is because a regular neighbourhood of $A \cup C$ (which is a disc) is just a 2-handle $H'$; the framing along which $H'$ is attached is determined by the intersection form, and is bound to be $n$.


Now the second claim is that $W := X\_1 \setminus {\rm Int\,} X\_0$ is an integral homology cobordism from $Y\_0 := S^3\_n(K\_0)$ to $Y\_1 := S^3\_n(K\_1)$. I will use excision, which tells us that $H\_i(W, Y) = H\_i(X\_1, X\_0)$ for each $i$.
Since $H\_i(X\_0) = H\_i(X\_1)$ is trivial when $i \neq 0,2$, and since at the level of $H\_0$ nothing really happens, we only need to look at $H\_2$.


Now, $H\_2(X\_0)$ is generated by a class represented by a Seifert surface for $K\_0$ capped with the core of the 2-handle, that is $A \cup C$. This surface intersects geometrically the co-core $D$ of the 2-handle $H$ of $X\_1$ once (since this intersection takes place in $H$, it's exactly $D\cap C$, which is one point), so the generator of $H\_2(X\_0)\simeq \mathbb Z$ is sent to a generator of $H\_2(X\_1) \simeq \mathbb Z$. It follows that the relative homology is trivial, as we wanted to show.


As for the addition: any integral homology cobordism invariant now gives a wealth of knot invariants. The Rokhlin invariant, for instance, gives you the concordance invariance of the Arf invariant. I am very partial to Heegaard Floer homology, so correction terms there give you a wealth of concordance invariants. (It should be pointed out that correction terms in Heegaard Floer homology were inspired by work of Frøyshov in Seiberg–Witten theory.)


