Let $Z\_1$ and $Z\_2$ be two closed subschemes of a smooth, complex, algebraic variety $X$. We will be interested in the Grothendieck ring of coherent sheaves of $X$, i.e. isomorphism classes of such sheaves modulo exact sequences and a product given by the tensor product of $\mathcal{O}\_X$-modules.


Denote by $[\mathcal{O}\_{Z\_1}]$ the class of $f^\ast\mathcal{O}\_{Z\_1}$ in the Grothendieck Ring of coherent sheaves on $X$, where $f:Z\_1\to X$ is the inclusion. 


Assume now that $Z\_2$ is a flat degeneration of $Z\_1$, i.e. there is a morphism $\pi: Y\to \mathbb{A}^1$ of schemes such that the generic fiber of $\pi$ is isomorphic to $Z\_1$ and the fiber of $\pi$ over $0$ is isomorphic to $Z\_2$.


I am reading a paper by David Speyer with the title 
*"A matroid invariant via the K-theory of the Grassmannian"* and on page 885, right below equation (3) he says that $[\mathcal{O}\_{Z\_1}]=[\mathcal{O}\_{Z\_2}]$ holds in this case. I don't see how to prove this implication and I'd be very grateful if someone could explain it to me. 



Why does a flat degeneration induce equality in the K-Theory?
As Allen says, we are talking about flat subfamilies of $X$. I.e. $Y$ is a closed subscheme of $X \times \mathbb{A}^1$, flat over $\mathbb{A}^1$, with $Z\_0 \times \{ 0 \}$ and $Z\_1 \times \{ 1 \}$ the fibers over $0$ and $1$.


Replacing $Y$ by its closure in $X \times \mathbb{P}^1$, we may assume that we instead have $Y$ closed in $X \times \mathbb{P}^1$, flat over $\mathbb{P}^1$. (Since $\mathbb{P}^1$ is smooth and one dimensional, being flat over it just means all associated primes of $Y$ map to the generic point of $\mathbb{P}^1$.) 


On $\mathbb{P}^1$, we have short exact sequences
$$0 \to \mathcal{O}(-1) \to \mathcal{O} \to k\_0 \to 0$$
and 
$$0 \to \mathcal{O}(-1) \to \mathcal{O} \to k\_1 \to 0$$
where $k\_0$ and $k\_1$ are the skyscraper sheaves of a point at $0$ and $1$ respectively. So the point $0$ and the point $1$ have the same class in $K\_0(\mathbb{P}^1)$. Since $Y \to X$ is flat, we can pull this back to an equality $[\mathcal{O}\_{Z\_0}] = [ \mathcal{O}\_{Z\_1}]$ on $Y$. 


Since $Y$ is closed in $X \times \mathbb{P}^1$, the map $\pi: Y \to X$ is proper and we can push forward this equality to $X$. We need to check that $\pi\_{\ast} \mathcal{O}\_{Z\_i \times \{ i \}} = \mathcal{O}\_{Z\_i}$ and $R^j \pi\_{\ast} \mathcal{O}\_{Z\_i \times \{ i \}} =0$ in order to know that the $K$-theoretic push forward is the obvious thing. 


I'm not sure where to cite this argument to, it is something that everyone I talked to about this sort of thing seemed to view as obvious. Allen does the analogue for equivariant $K\_0$, which is more interesting, in Prop 3.5 of his [recent paper](http://arxiv.org/abs/1408.1261).


