Consider a cartesian closed category $\mathcal E$. Let $1$ be the terminal object and $U, V\hookrightarrow 1$ to monomorphisms into $1$. I wanted to prove that $U^V\rightarrow 1$ is also a monomorphism. So, we want to prove that for every $X$ there is at most one map $X\rightarrow U^V$. This should be immediate maps $X\rightarrow U^V$ are in 1-1, onto correspondence with maps $X\times V\rightarrow U$, but $U\rightarrow 1$ is mono, so there is at most one map $X\times V\rightarrow U$.

Now, I didn't use that $V\rightarrow 1$ is mono, so I would conclude that this hypothesis is superfluous. Is my conclusion correct? For reference, I'm working out the statement in "Sheaves in Geometry and Logic", page 201.

The exponential of $U^V$ of two open objects $U\hookrightarrow 1$ and $V\hookrightarrow 1$ is again open, i.e., $U^V\rightarrow 1$ is monic

  • 3
    $\begingroup$ Yes, $V$ may be arbitrary. $\endgroup$
    – Zhen Lin
    Dec 30, 2021 at 12:21
  • $\begingroup$ Thanks. Do you want to make it into an answer? So I can accept it $\endgroup$
    – Alessandro
    Dec 30, 2021 at 13:08

1 Answer 1


Yes, $V$ may be arbitrary. So long as $U$ is subterminal $U^V$ will also be subterminal. In other words, you have shown that the class of subterminal objects is an exponential ideal.


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