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Let $C$ be a concrete category, let $F$ be an object in $C$, let $X$ be a nonempty set, let $f : X → F$ be a function between sets. Let $D$ be a category whose objects are pairs $(B, g)$, where $g : X → B$ are functions between sets, and morphisms from the class $Hom((B, g),(H, h))$ are defined as morphisms $ φ: B → H$ in $C$ such that $φ \circ g = h.$ Show that $F$ is a free object with basis $X$ in $C$ if and only if $(F, f)$ is an initial object in $D.$

Definition of free object: Let $C$ be a concrete category, let $F$ be an object in $C$, let $X$ be a nonempty set, let $f : X → F$ be a function between sets. The object $F$ is called free with basis $X$ if and only if for each object $H$ and each function between sets $h : X → H$ there is exactly one morphism $φ:F→H$ such that $φ \circ f = h.$

Definition of initial object: Let $C$ be a category. An object $I$ of $C$ is called an initial object, if for each object $T$ of $C$ there is exactly one morphism $i:I→T.$

So, my question: is it as easy as I think? I mean both ways, there is nothing to check because this is straight from definition.

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  • $\begingroup$ Yes, there isn't much to check other than the definitions. I don't see why the author desires to exclude free objects with basis $\emptyset$, since not only there are notable examples, but it all works exactly the same. $\endgroup$ – Saucy O'Path May 11 at 16:41

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