0
$\begingroup$

According to this definition of an essentially surjective functor:

[A functor] $F: C \rightarrow D$ is essentially surjective if for every object $y$ of $D$, there exists an object $x$ of $C$ and an isomorphism $F(x) \cong y$ in $D$.

Question: What is here meant by the $F(x) \cong y$? Does it mean that there exists an arrow $h$ in $D$ s.t. $h: F(x) \rightarrow y$ is invertible in $D$ (i.e., that $h$ is an isomorphism in $D$)?

  • 1
    It means $F(x)$ and $y$ are isomorphic, yes. – 2017-01-31
  • 0
    So my articulation using $h$ above is correct? – 2017-01-31
  • 2
    Yes, it is. ${}{}$ – 2017-01-31

0 Answers 0