[In this post "ring" means "commutative ring with one".]
Let $A$ be a noetherian ring, and let $f:A\to A$ be an endomorphism which is also an epimorphism.
Is $f$ necessarily injective?
Eric Wofsey gave an example of an epimorphic endomorphism of a noetherian ring which is not surjective. (It is well known, and easy to prove, that surjective endomorphisms of noetherian rings are automatically bijective.)
[By definition, a morphism of rings $f:A\to B$ is an epimorphism if for all pairs of morphisms $(g,h):B\rightrightarrows C$ the equality $g\circ f=h\circ f$ implies $g=h$. Surjective morphisms are epimorphic, but the converse does not always hold: for instance the inclusion $\mathbb Z\to\mathbb Q$ is an epimorphism.]
For more details about epimorphisms see
$\bullet$ MathOverflow thread What do epimorphisms of (commutative) rings look like?.
$\bullet$ Stacks Project Section Epimorphisms of rings.
$\bullet$ Samuel Seminar .
$\bullet$ MathOverflow thread Is Krull dimension non-increasing along ring epimorphisms?.