Let $(\mathcal{C},F)$ be a concrete category (so that the functor $F: \mathcal{C} \to \mathbf{Set}$ is faithful). Let $A$ be a set (called basis), let $X$ be an object of $\mathcal{C}$, and let $i: A \to F(X)$ be a function (called canonical injection). Then $X$ is called a free object (with respect to $i$) in a category $\mathcal{C}$ on a set $A$ if for any object $Y$ of $\mathcal{C}$ and for any function $f: A \to F(Y)$ there is a unique morphism $g: X \to Y$ so that $F(g) \circ i = f$.

Now, I need to know how is it possible to find out that $i$ is an injection (that is, a monomorphism in a category of sets).

  • $\begingroup$ Not unrelated: math.stackexchange.com/a/1219928/166694 $\endgroup$ – Stefan Perko Nov 13 '16 at 22:13
  • $\begingroup$ @Derek I haven't made any progress. I thought if we could get such object $Y$ of $C$ so that $F(Y) = A$, then we could get a morphism $\phi: X \to Y$ so that $F(\phi): F(X) \to F(Y) = A$ is a left-inverse of $i$. But we don't know such an object exists. $\endgroup$ – Jxt921 Nov 14 '16 at 2:04
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    $\begingroup$ @Jxt921 Okay. Perhaps try using the more direct notion of monomorphism (or injective function), namely $i$ is a mono if $i \circ h = i \circ k$ implies $h = k$. Now, categorically this needs to be for all possible $h$ and $k$ that compose with $i$, but if we restrict to $h, k : A \to A$ we can apply the definition of free object. To generalize this to get what we want, we now need to use properties of $\mathbf{Set}$. $\endgroup$ – Derek Elkins Nov 14 '16 at 2:49
  • $\begingroup$ @DerekElkins Thanks for the hint, but, unfortunately, as of now, I couldn't get it. I don't understand how we can get a first step. Assume $i \circ h = i \circ k, \ h, k: A \to A$. So, do you propose that we "put" functions $i \circ h$ and $i \circ k$ into the diagram and get two unique morphisms $\phi, \psi: X \to X$ so that $F( \phi ) \circ i = i \circ h$ and $F( \psi ) \circ i = i \circ k$? Yeah, since $i \circ h = i \circ k$ we have $\phi = \psi$, but what it gives us? But, probably, I misunderstood you. I would be grateful if you can expand on that matter. $\endgroup$ – Jxt921 Nov 14 '16 at 4:34

My comment may not have been as helpful as I hoped. Here's a brute-force answer.

Assume $a, b \in A$ and $a \neq b$. (If $A$ has 0 or 1 element, then $i$ is trivially injective.) Assume $x \in FX$ and $x \neq i(a)$. (This leaves the case where $FX$ has only one element but $A$ has more than one open.) Now, assume for contradiction that $i(a) = i(b)$. Define $$h(z) = \begin{cases}x,& z = a \\i(z),& z \neq a\end{cases}$$ Let $\varphi$ witness the universal property, i.e. $F\varphi(f)\circ i = f$. Then $$h(a) = (F\varphi(h))(i(a)) = (F\varphi(h))(i(b)) = h(b)$$ which contradicts $h(a) = x \neq i(a) = i(b) = h(b)$.

If $|FY|>1$ for any $Y$, you can immediately derive a contradiction to the universal property if $|A| > 1$. The only remaining case is $|FY| = 1$ for all $Y$. Particularly, the case when $C$ is the terminal category. This is a counter-example. The only function $A \to FY$ for any $Y$ is $i$. You can check that it satisfies the universal property for any $A \neq \emptyset$. $i$ will only be injective if $|A| = 1$.


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