0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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 $|A| = 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.