# Arrows from initial objects to non-isomorphic objects in an elementary topos are monomorphisms?

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.

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