A decision problem $C$ is $NP$-complete if $C$ is in $NP$, and every problem in $NP$ is reducible to $C$ in polynomial time.
Reduction means transforming an instance of one problem $A$ to an instance of another problem $B$, and by using an algorithm that solves $B$ we can obtain the solution to $A$ for our original instance.
Is there any relationship between the size of instance before and after reduction? If for example the algorithm solving $A$ is $O(n^2)$ and our instance of $A$ of size $n$, what can we say about the instance size after reducing it to $B$, whose algorithm is $O(2^n)$ (the complexity of algorithms actually doesn't matter). Is it $n$ as well? Or maybe it is a polynomial of $n$?
The question is if the fact that a problem $A$ can be reduced to $B$ in polynomial time tells us anything about the size of the resulting instance (of problem $B$) with respect to the size of initial instance of $A$.
Why? It is known that
If we had a polynomial algorithm for an $NP$-complete problem we could solve any other $NP$ problem in polynomial time
right? Because we just reduce $NP$ problem to $NP$-complete problem in polynomial time and solve it in polynomial time.
Let's say problem $A$ is $NP$ and problem $B$ is $NP$-complete. I've found an algorithm for $B$ that is $O(n)$ (polynomial). And let's say the size of our $A$ instance is $k$. If I reduced it to $B$ in polynomial time and got an instance of $B$ of size $n=2^k$, then the total time of solving it wouldn't be polynomial. So certainly the polynomial time reduction can't give an instance of size $2^k$, because if it could, then the quote above wouldn't be true.