Pulling back the "test-element"? (Orthogonal factorization systems)

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Below is the statement and proof of Lemma 3.1 from the paper Reflective subcategories, localizations, and factorization systems by Cassidy Hébert and Kelly.

Lemma 3.1. Let $\mathcal N$ be a class of monomorphisms in $\mathsf C$, closed under composition; let $\mathsf C$ admit all pulbacks (along any map) of maps in $\mathcal N$, and all intersections of maps in $\mathcal N$; and let these pullbacks and intersections again belong to $\mathcal N$. Then $(\mathcal N^\perp,\mathcal N)$ is a strong factorization system.

Proof. $f:A\to B$ factors as $me$, where $m$ is the intersection of the $\mathcal N$-subobjects of $B$ through which $f$ factors; it is easy to see that $e\in \mathcal N^\perp$, by pulling back the "test-element" of $\mathcal N$.

What is meant by "test-element"?

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To prove that $e : A \to f(A) \in \mathcal{N}^\bot$, you have to pick arbitrary $m \in \mathcal{N}$ and show that if $$ \require{AMScd} \begin{CD} A @>{e}>> f(A)\\ @VVV @VVV \\ C @>{m}>> D \end{CD} $$ commutes, then there exists a diagonal $f(A) \to C$. Presumably, the "test element" refers to $m$, because the proof goes as follows: take the pullback of $f(A) \to D \leftarrow C$; then, since $e$ factorizes through this pullback, the projection onto $f(A)$ must be an isomorphism. Inverting this isomorphism and composing it with the projection onto $C$ gives the desired morphism $f(A) \to C$.