Seifert surfaces of knots Let $K$ be a null-homologous knot in a $3$-manifold $M$, i.e., $[K] = 0$ in $H_1(M)$. In this post, it was elegantly shown that there is a Seifert surface $F \subset M$ such that $\partial F =K$.
I wonder if the converse statement is also true. If we have such a Seifert surface, can we guarantee that the knot $K$ must be null-homologous inside $M$?
P.S. Any reading advice also would be nice.
 A: First, to review the meaning of $[K]$, for any oriented knot $K$ in any topological space $X$, there is a corresponding homology class in $H_1(X)$ that I'll denote $[K]_X$, and it is defined like this: Choose an orientation preserving homeomorphism $f : S^1 \to K$, consider the induced homology homomorphism
$$\mathbb Z = H_1(S^1) \xrightarrow{f_*} H_1(K) \xrightarrow{i_*} H_1(X)
$$
where $i$ is the inclusion map, and define
$$[K]_X = (i_* \circ f_*)(1) = (i \circ f)_*(1)
$$
Now, perhaps you have done the algebraic topology exercise to prove that if $F$ is a compact, oriented surface and $K$ is its boundary circle then $[K]_F=0$.
This exercise applies in your situation, because a Seifert surface of $K$ in $M$ is just the inclusion $F \xrightarrow{j} M$ of some compact, oriented surface $F$ in $M$ such that $K$ is the boundary of $F$.
So to answer your question, consider the composition
$$S^1 \xrightarrow{f} K \xrightarrow{i} F \xrightarrow{j} M
$$
From the algebraic topology exercise it follows that $[K]_F=0$. Since $j \circ i : K \to M$ is simply the inclusion map of $K$ into $M$, it follows that
$$[K]_M = ((j \circ i) \circ f)_*(1) = j_*((i \circ f)_*(1)) = j_*([K]_F) = j_*(0)=0
$$
The second equals sign is a simple application of the functorial properites of induced homology homomorphisms.
