How to show that a problem is easy?

Let $P$ be a problem that you need to study its difficulty, i.e., NP-hard, Polynomial-time solvable, etc.


My question is: If I reduce a known polynomial time problem (say,
maximum matching in bipartite graph) to $P$, why I can say that $P$ is an
easy problem?


My guess is: No, we cannot say that.


Why? Because from an instance of maximum matching problem, $I_{ MM }$,
I create an instance of $P$, $I_{ P }$, and then I show that maximum
matching is solved with $I_{ MM }\iff P$ is solved with $I_{ P }$.


But what if from another instance of maximum matching problem, $I_{ MM }'$
, I create another instance of $P$, $I_{ P }'$, which is hard to
solve?

I have read that the reduction is correct and works, for example from sorting to convex hull, but I do not why.

I do not know what I am missing here.

You are right. Any problem in $\mathsf{P}$ can be (polytime) reduced to the halting problem, for example. (In fact, any problem in $\mathsf{P}$ can be polytime reduced to any non-trivial language, that is any language other than $\emptyset$ or $\Sigma^*$.)

What is true is that if A reduces to B and B is easy, then so is A. In particular, if A polytime reduces to B and B is in $\mathsf{P}$, then A is also in $\mathsf{P}$.

StackExchange Computer Science Q#52275, answer score: 3