Let $A,B$ be sets, and $A\sqcup B$ the disjoint union. Suppose that in a (concrete) category, the free objects $F_A,F_B,F_{A\sqcup B}$ exist, and that the coproduct $F_A \coprod F_B$ exists. How can I prove (if true) that $$F_{A\sqcup B}\;\cong\;F_A\coprod F_B?$$

Using the universal properties, I've managed to produce a unique morphism $f\!: F_A\coprod F_B \longrightarrow F_{A\sqcup B}$, such that the following diagram commutes: enter image description here

Now I don't know how to produce a morphism $g\!: F_{A\sqcup B} \longrightarrow F_A\coprod F_B$.

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    $\begingroup$ If $F : C \to C'$ is any left adjoint, then $F$ preserves colimits. $\endgroup$ – Martin Brandenburg Jan 4 '13 at 15:27
  • $\begingroup$ Did you manage to solve this exercise now or do you need more indications? $\endgroup$ – Martin Jan 5 '13 at 6:41
  • $\begingroup$ @Martin: Apologies for the delayed answer. I am not familiar with adjoint functors. In mland's answer, he mentions an adjoint functor. Is that meant the functor $\mathcal{C}\to \mathrm{Set}$ from the definition of a concrete category? In the formulation of the question, I didn't assume that coproducts and free objects always exist, just that $F_A,F_B,F_{A\sqcup B}, F_A\coprod F_B$ do. So $F_-$ need not be always defined, hence not a functor. I thought there would be some easy way to find $g\!:F_{A\sqcup B}\to F_A\coprod F_B$ 'by hand'. $\endgroup$ – Leo Jan 5 '13 at 18:44
  • $\begingroup$ Yes, Martin B. and mland assume that the functor $U \colon \mathcal{C} \to \mathsf{Set}$ has a left adjoint $F \colon \mathsf{Set} \to \mathcal{C}$. This means exactly that $F(A)$ is free over $A$ in your sense. A concrete example where U fails to have a left adjoint is to take the category of finitely generated modules over a ring. $F_A$ is defined for $A$ finite, but not for $A$ infinite. $\endgroup$ – Martin Jan 6 '13 at 0:22
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    $\begingroup$ Of course the same argument works when $F$ doesn't exist globally. See also the answer by User 24601. $\endgroup$ – Martin Brandenburg Jan 6 '13 at 4:12

You mentioned that you don't know about adjoint functors, so probably this is as good a time as any to learn about them. Since the formulation of your question does not assume the existence of a left adjoint $F$ to the "forgetful functor" $U \colon \mathcal{C} \to \mathsf{Set}$, we shouldn't do what Martin B. and mland suggested, although the argument is essentially the same.

I gave the construction of the map $g \colon F_{A \amalg B} \to F_A \amalg F_B$ in a comment to mland's answer which I describe in more detail in the highlighted box below. Let me try to describe the thought process that led to it.

  1. What we are given:

    • a concrete category $\mathcal{C}$ with "forgetful functor" $U \colon \mathcal{C} \to \mathsf{Set}$;

    • sets $A,B$ and a coproduct diagram $j_A \colon A \to A \amalg B \gets B : \! j_B$ in $\mathsf{Set}$;

    • objects $F_A, F_B, F_{A \amalg B}$ in $\mathcal{C}$ and functions $\eta_A \colon A \to UF_A$, $\eta_B \colon B \to UF_B$ and $\eta_{A \amalg B}\colon A \amalg B \to UF_{A \amalg B}$ in $\mathsf{Set}$ exhibiting $F_A,F_B, F_{A \amalg B}$ as free objects over $A,B,A \amalg B$, respectively;

    • a coproduct diagram $i_A \colon F_A \to F_{A} \amalg F_{B} \gets F_B : \! i_B$ in $\mathcal{C}$.

  2. What you did:

    • The universal property of $\eta_{A} \colon A \to UF_A$ applied to $\eta_{A \amalg B} \circ j_A \colon A \to UF_{A\amalg B}$ gives a unique map $f_{A} \colon F_A \to F_{A \amalg B}$ such that $Uf_A \circ \eta_A = \eta_{A \amalg B} \circ j_A$ and similarly there is a unique map $f_B \colon F_B \to F_{A \amalg B}$ such that $Uf_B \circ \eta_B = \eta_{A \amalg B} \circ j_B$.

    • From the coproduct diagram $i_A \colon F_A \to F_{A} \amalg F_{B} \gets F_B : \! i_B$ and the diagram $f_A \colon F_A \to F_{A \amalg B} \gets F_{B} : \! f_B$ we get a unique morphism $f \colon F_A \amalg F_B \to F_{A \amalg B}$ such that $f_A = f \circ i_A$ and $f_B = f \circ i_B$.

  3. What is missing: a map $g \colon F_{A \amalg B} \to F_A \amalg F_{B}$ inverse to $f$.

  4. What we haven't used so far:

    • the universal property of the map $\eta_{A \amalg B} \colon A \amalg B \to UF_{A \amalg B}$;

    • the fact that $j_A \colon A \to A \amalg B \gets B : \! j_B$ is a coproduct diagram in $\mathsf{Set}$.

So the solution likely involves combining the two missing points. The universal property of $\eta_{A \amalg B}$ will give us a map defined on $F_{A \amalg B}$ which will land in $X$ whenever we produce a map $A \amalg B \to UX$ for $X$ in $\mathcal{C}$ and the coproduct $A \amalg B$ is designed to produce maps out of $A \amalg B$, so we want to take $X = F_A \amalg F_B$ and we need to produce maps $A \to U(F_A \amalg F_B) \gets B$.

The forgetful functor $U$ applied to the coproduct diagram $$i_A \colon F_A \to F_{A} \amalg F_{B} \gets F_B : \! i_B$$ yields a diagram $$Ui_A \colon UF_A \to U(F_{A} \amalg F_{B}) \gets UF_B : \! Ui_B.$$ Moreover, we are given $\eta_A\colon A\to UF_A$ and $\eta_B \colon B \to UF_B$. Thus, we get the diagram $$h_A = Ui_A \circ \eta_A \colon A \to U(F_{A} \amalg F_{B}) \gets U(F_B) : \! h_B = Ui_B \circ \eta_B$$ and therefore a unique map $h \colon A \amalg B \to U(F_{A} \amalg F_{B})$ such that $h\circ j_A = h_A$ and $h \circ j_B = h_B$.

From universality of $\eta_{A \amalg B} \colon A \amalg B \to UF_{A \amalg B}$ there is a unique $g \colon F_{A \amalg B} \to F_{A} \amalg F_{B}$ such that $Ug \circ \eta_{A \amalg B} = h$.

It remains to verify that $f \circ g \colon F_{A \amalg B} \to F_{A \amalg B}$ and $g \circ f \colon F_{A} \amalg F_{B} \to F_{A} \amalg F_{B}$ are the identity morphisms by using the given definitions and universal properties. This is a routine exercise that tends to be clearer if you do it yourself, so I recommend that you try to do it on your own before reading the details.

Consider $f \circ g$. We apply the universal property of $\eta_{A \amalg B}\colon A \amalg B \to F_{A \amalg B}$. First, we have $$ U(f \circ g) \circ \eta_{A \amalg B} = Uf \circ Ug \circ \eta_{A \amalg B} = Uf \circ h. $$ But $h$ is determined by $h \circ j_A = h_A = Ui_A \circ \eta_A$, so $$ Uf \circ h \circ j_A = Uf \circ Ui_A \circ \eta_A = U(f \circ i_A) \circ \eta_A = U(f_A) \circ \eta_A = \eta_{A \amalg B} \circ j_A, $$ whence $Uf \circ h \circ j_A = \eta_{A \amalg B} \circ j_A$. Similarly $Uf \circ h \circ j_B = \eta_{A \amalg B} \circ j_B$. Therefore $Uf \circ h = \eta_{A \amalg B}$ and hence $U(f \circ g) \circ \eta_{A \amalg B} = \eta_{A \amalg B} = U(1_{F_{A \amalg B}}) \circ \eta_{A \amalg B}$. This implies $f \circ g = 1_{F_{A \amalg B}}$ by universality of $\eta_{A \amalg B}$.

Now consider $g \circ f \colon F_{A} \amalg F_{B} \to F_{A} \amalg F_{B}$. We apply the universal property of $F_{A} \amalg F_{B}$. So we consider $g \circ f \circ i_A \colon F_{A} \to F_{A} \amalg F_{B}$ and we want to show that $g \circ f \circ i_A = i_A$. To do this, note that $$ U(g \circ f \circ i_A) \circ \eta_A = Ug \circ Uf_A \circ \eta_A = Ug \circ \eta_{A \amalg B} \circ j_A = h \circ j_A = h_A = Ui_A \circ \eta_A $$ so $g \circ f \circ i_A = i_A$ by universality of $\eta_A$ and thus $g \circ f \circ i_A = 1_{F_A \amalg F_B} \circ i_A$. Similarly, $g \circ f \circ i_B = 1_{F_A \amalg F_B} \circ i_B$, so the universal property of $F_A \amalg F_B$ yields $g \circ f = 1_{F_A \amalg F_B}$.

  • $\begingroup$ This is the answer I was looking for, though it is more complicated than I expected, with the multiple usages of universal properties. The main thing I was missing initially is that $\sqcup$ is a coproduct in $\mathsf{Set}$. Thank you! $\endgroup$ – Leo Jan 6 '13 at 7:03
  • $\begingroup$ Fixed the typo. Yes, it looks complicated but that's just the way it is. Every appeal to universality corresponds to one isomorphism in the answer with Yoneda. $\endgroup$ – Martin Jan 6 '13 at 7:55

You can produce the map you want as follows: Let me call the category $\mathcal{C}$ and denote the homsets by $\mathcal{C}(X,Y)$ for objects $X$ and $Y$.

A free functor just says it is a left adjoint to the forgetful functor $U$ to sets. Now by adjunction you have to give maps $A \to U(F_A\coprod F_B)$ and $B \to U(F_A\coprod F_B)$ because $A\cup B$ (which of course should be a disjoint union of sets) is a coproduct of $A$ and $B$ in sets. But you have canonical maps $A \to U(F_A)$ and $B \to U(F_B)$ given by the adjunction. Moreover as part of the coproduct in $\mathcal{C}$ there are maps $F_A \to F_A\coprod F_B$ and $F_B \to F_A\coprod F_B$. By functoriality you get maps $$ A \to U(F_A) \to U(F_A\coprod F_B) \text{ and } B \to U(F_B) \to U(F_A\coprod F_B).$$

But I would prove your claim differently. Every left adjoint commutes with colimits, so you can just by hands show that $F_{A\cup B}$ has the universal property of a coproduct, by showing that maps out of $F_{A\cup B}$ are products of maps out of $F_A$ and $F_B$. This again you do by the adjunction.

  • $\begingroup$ The formulation of the question assumes much less than the existence of a left adjoint to $U$. This doesn't change the essence of the argument but requires some minor changes in the write-up. $\endgroup$ – Martin Jan 4 '13 at 15:53
  • $\begingroup$ Yes, this is true. But as you said, this should be not to much work.. I will leave it this way (it at least solves the issue for categories where the forgetful functor admits a left adjoint) unless Leon wants me to change it. $\endgroup$ – mland Jan 4 '13 at 17:38
  • $\begingroup$ I didn't mean to say you should rewrite it. I pointed it out because the question appears to be carefully phrased so as to avoid this assumption. The small missing thing is that from the two maps in the display you get a map $h\colon A \cup B \to U(F_A \amalg F_B)$ and by the universality of $i_{A \cup B} \colon A \cup B \to UF_{A \cup B}$ there is a unique map $g \colon F_{A \cup B} \to F_A \amalg F_B$ such that $Ug \circ i_{A \cup B} = h$. This $g$ is the desired inverse to the map $f \colon F_{A} \amalg F_{B} \to F_{A \cup B}$ already constructed by Leon. $\endgroup$ – Martin Jan 4 '13 at 18:34
  • $\begingroup$ Perfectly right. Thanks for the correction. $\endgroup$ – mland Jan 4 '13 at 19:36

There's a much simpler proof using Yoneda.

By the definition of coproducts and free objects, there is a chain of isomorphism of sets

$$ \hom(F_A \coprod F_B, X) \cong \hom(F_A, X) \times \hom(F_B,X) \cong \hom_{set}(A,X_{set}) \times \hom_{set}(B,X_{set}) $$ $$ \cong \hom_{set}(A \coprod_{set} B, X_{set}) \cong \hom(F_{A \coprod_{set} B},X) $$ which is obviously (if tediously) natural in the variable $X$. In other words, the objects $F_A \coprod F_B$ and $F_{A \coprod_{set} B}$ co-represent the same functor $\mathcal C \to Sets$. By the Yoneda Lemma on $\mathcal{C}^{op}$, the two objects are isomorphic.

(A note on notation in case you need it: Above, $\hom = \hom_{\mathcal{C}}$ in the concrete category $\mathcal{C}$, and $\hom_{set}$ is the hom in the category of sets. $\coprod_{set}$ indicates the coproduct (disjoint union) in sets. Given an object $X$ in the concrete category, $X_{set}$ is the object one obtains by taking the functor from the concrete category to the category of sets.)

If you're starting to learn this stuff, this proof is probably more appealing. But as you go onto learn more, I definitely recommend the (more powerful) chain of thought: "Free's a left adjoint so it preserves all colimits."

Finally, sorry that I don't explicitly exhibit the inverse you seek. But I didn't feel like tracing through Yoneda and all that.

  • $\begingroup$ I am also not familiar with the Yoneda lemma, but it looks quite magical. One day your answer will prove valuable. Thanks! Is there an example of a category in which a free object on a set doesn't exist, and a coproduct of two objects doesn't exist? Is this a rare occurence? $\endgroup$ – Leo Jan 6 '13 at 6:55
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    $\begingroup$ The rarity depends on where you work. Most natural categories in algebra (modules, rings, groups) have a concrete functor to sets and admit a left adjoint called "Free." An example of a concrete category that doesn't (among many you can cook up): Let P be a set. This defines a category $\mathcal{P}$ whose objects are subsets $S \subset P$ and whose morphisms $hom(S,S')$ are empty, or a single point if $S \subset S'$. (If you like, its morphisms are the inclusions -- more generally, any poset yields a category.) The functor $\mathcal{P} \to Sets$ sending $S \mapsto S$ has no left adjoint. $\endgroup$ – user54535 Jan 6 '13 at 18:02

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