Let $\mathscr{T}$ be an elementary topos (I use the definition of a Cartesian closed, finitely complete category with a subobject classifier). Let $a$ be any object (that's not isomorphic to $0$) of $\mathscr{T}$, and $0$ the initial object. Is any arrow of the form $f: 0 \to a$ a monomorphism? I believe the answer is yes, but I can't figure out a proof.

  • $\begingroup$ Possible duplicate of The arrows from the initial object in a category are monomorphisms? $\endgroup$ – Victoria M Feb 1 at 20:26
  • 1
    $\begingroup$ No my question is targeted towards elementary toposes, not all categories in general. $\endgroup$ – SpooFwen Feb 1 at 20:27
  • $\begingroup$ Elementary topoi are just cartesian closed categories with both finite limits and a subobject classifier. $\endgroup$ – Victoria M Feb 1 at 20:29
  • 1
    $\begingroup$ @VictoriaM : yes but the answer is different for general categories and for topoi $\endgroup$ – Max Feb 1 at 20:58
  • $\begingroup$ @Max : Ah I see, my apologies $\endgroup$ – Victoria M Feb 1 at 21:00

The answer is that they are, but for stupid reasons : the same reason that in $\mathbf{Set}$, any map $\emptyset \to A$ is mono.

Indeed, in an elementary topos $\mathscr{T}$, any object $A$ with a map $A\to 0$ is also $0$, so this map is actually unique.

To prove this, first note that for any object $B$, $B\times 0 \simeq 0$. Indeed, by exponentiation, one has $\hom(B\times 0, X)\cong \hom(0,X^B) \cong \{*\}$, naturally in $X$, so $B\times 0$ is an initial object.

Then note that if you have a map $f:A\to 0$, then you have a map $A\to A\times 0$, namely $(id_A,f)$. Now $\pi_A\circ (id_A,f) = id_A$ by definition, and $\pi_A\circ (id_A,f)\circ \pi_A = id_A\circ \pi_A = \pi_A=\pi_A\circ id_{A\times 0}$, $\pi_0\circ (id_A,f)\circ \pi_A = f\circ \pi_A= \pi_0\circ id_{A\times 0}$ because $A\times 0$ is initial, and so there's a unique map $A\times 0 \to 0$ : if I know one such map, it must be this one.

These two equations prove (universal property of the product) that $(id_A,f)\circ \pi_A = id_{A\times 0}$; so that $(id_A,f)$ and $\pi_A$ are a pair of isomorphisms : $A\simeq A\times 0$. But $A\times 0 \simeq 0$, so $A\simeq 0$.

Now we proved that any object with a map $A\to 0$ must be initial, so if there are two maps $f,g: B\to 0$ such that [insert any condition you like], then $f=g$. In particular, any map leaving $0$ is a monomorphism


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.