Showing some object is initial This is a problem from logic. I'll first phrase it in general categorical terms since it is rather obscure, and then with context.
Let $\mathcal{E}$ be a regular category where all epi's are regular. If $f: V \to X$ factors through $\bot \to X$ where the latter morphism is the mono part of the factorization $0 \to \bot \to X$, is then $V$ initial?
Context
I am trying to prove iv of the soundness of Kripke-Joyal semantics. By iii and the definition $\neg N = N \implies \bot$ where $\implies$ is the Heyting implication and $\bot$ is the bottom element in the subobject poset of $X$, it is enough to show that for $V \xrightarrow{g} U \xrightarrow{\alpha} X$ we have that $g$ factors through $\bot \to X$ if and only if $V$ is initial.
 A: The answer to the question you asked is no. For one thing, not every regular category has an initial object. But putting that issue aside, the category of groups is a regular category with an initial object in which all epis are regular. For any groups $G$ and $H$, we have the trivial homomorphism $f\colon G\to H$ which factors through the trivial group as $G\to 0 \to H$. This does not imply that $G$ is initial (trivial). 
On the other hand, as Mark Kamsma suggests in the comments, this is true in any coherent category, and therefore in any elementary topos. 
Recall that a coherent category is a regular category such that for every object $X$, the poset $\text{Sub}(X)$ has finite joins (including the empty join $\bot_X$) and these joins are preserved by pullbacks. Now any coherent category has a strict initial object $0$, and $0 = \bot_1$, the smallest subobject of the terminal object (see Lemma A1.4.1 in the Elephant). 
Since the empty join is stable under pullback, we can pull back the arrow $0 = \bot_1 \to 1$ along the unique arrow $X\to 1$ to obtain $\bot_X\to X$. But then $\bot_X$ admits an arrow to $0$ (one side of the pullback square), and since $0$ is strict initial, $\bot_X\cong 0$. 
Now for any $V\to X$ which factors through $\bot_X\to X$, $V$ admits an arrow to $0$, so also $V\cong 0$ and $V$ is initial.
