Monomorphisms, Injectivity, and the Free-Forgetful Adjunction in Concrete Category over Set

IntroductionFree-forgetful adjunction in $\mathrm{Top}, \mathrm{Mon}, \mathrm{Grp}, \mathrm{Ab}, \mathrm{Ring}, R\text{-}\mathrm{Mod}$...

Introduction

When we study $\mathrm{Ring}$$\mathrm{Ring}$, we see that epimorphisms are not equivalent to surjections. For example, let $R$$R$ be an integral domain and $F(R)$$F(R)$ be its field of fractions (or more generally, any localization). Then the canonical map $\iota :R\hookrightarrow F(R)$$\iota: R \hookrightarrow F(R)$ is an epimorphism but not a surjection. Another example is consider a full subcategory of monoid with $\mathrm{Ob}=\{({\mathbb{N}}^{+},+),\{1\}\}$$\mathrm{Ob}=\{(\mathbb N^+,+),\{1\}\}$, denote it by $\mathcal{N}$$\mathcal N$ . Then it is a concrete category over $\mathrm{Set}$$\mathrm{Set}$. All the morphism are mono since $ab=ac⟹b=c$$ab=ac\implies b=c$. Since it is commutative we have all the morphism are epi as well. But the only surjections are $\mathrm{id}$$\mathrm{id}$ on each object. For $\mathrm{Top}$$\mathrm{Top}$, let $Y$$Y$ be a Hausdorrf space and $i:X\to Y$$i:X\to Y$ be a dense subspace. Then $i$$i$ is an epi morphism. Since for $f,g:Y\to Z$$f,g:Y\to Z$, $f{|}_X=g{|}_X$$f|_X=g|_X$, then $X\subseteq \mathrm{Eq}(f,g)=(f,g{)}^{-1}(\mathrm{\Delta }(Y))$$X\subseteq\mathrm{Eq}(f,g)=(f,g)^{-1}(\Delta(Y))$, where $\mathrm{\Delta }(Y)=\{(y,y)|y\in Y\}$$\Delta(Y)=\{(y,y)|y\in Y\}$. By $Y$$Y$ is Hausdorff, $\mathrm{\Delta }(Y)$$\Delta(Y)$ is closed, hence $\mathrm{Eq}(f,g)$$\mathrm{Eq}(f,g)$ is a closed set contain a dense subset $X$$X$. Hence $\mathrm{Eq}(f,g)$$\mathrm{Eq}(f,g)$ is $Y$$Y$, hence $f=g$$f=g$, i.e. $i$$i$ is an epi morphism but $i$$i$ is not surjective.

However, in many concrete categories over $\mathrm{Set}$$\mathrm{Set}$, such as $\mathrm{Top},\mathrm{Mon},\mathrm{Grp},\mathrm{Ring},R\text{-}\mathrm{Mod}$$\mathrm{Top}, \mathrm{Mon}, \mathrm{Grp}, \mathrm{Ring}, R\text{-}\mathrm{Mod}$, we find that a morphism $f$$f$ is a monomorphism if and only if $f$$f$ is injective.

Is there any reason why $f$$f$ being a monomorphism is equivalent to $f$$f$ being injective? Why does this equivalence hold for monomorphisms but not for epimorphisms? What breaks the expected duality between these concepts?

The reason is: the free functor is left adjoint to the forgetful functor!

---

Free-forgetful adjunction in $\mathrm{Top},\mathrm{Mon},\mathrm{Grp},\mathrm{Ab},\mathrm{Ring},R\text{-}\mathrm{Mod}$$\mathrm{Top}, \mathrm{Mon}, \mathrm{Grp}, \mathrm{Ab}, \mathrm{Ring}, R\text{-}\mathrm{Mod}$...

In the categories mentioned above, each is a concrete category over $\mathrm{Set}$$\mathrm{Set}$—that is, there exists a faithful functor from the category to $\mathrm{Set}$$\mathrm{Set}$. The forgetful functor admits a left adjoint, called the free functor.

For example:

• $\mathrm{Top}$$\mathrm{Top}$ → discrete topology functor

• $\mathrm{Mon}$$\mathrm{Mon}$ → free monoid functor

• $\mathrm{Grp}$$\mathrm{Grp}$ → free group functor

• $\mathrm{Ring}$$\mathrm{Ring}$ → polynomial ring functor

• $R\text{-}\mathrm{Mod}$$R\text{-}\mathrm{Mod}$ → free $R$$R$-module functor

Let us denote the forgetful functor by $U$$U$ and the free functor by $F$$F$. Then the free-forgetful adjunction gives the following natural isomorphism:

$${\mathrm{Hom}}_{\mathcal{C}}(F(X),Y)\cong {\mathrm{Hom}}_{\mathrm{Set}}(X,U(Y))$$

In particular, we have:

$${\mathrm{Hom}}_{\mathcal{C}}(F([1]),Y)\cong {\mathrm{Hom}}_{\mathrm{Set}}([1],UY)\cong UY$$

Remark. The key point here is the identity functor of $\mathrm{Set}$$\mathrm{Set}$ is representable.

Hence, the forgetful functor $U$$U$ is representable via $F([1])$$F([1])$, where $[1]$$[1]$ is the singleton set.

The free object corresponding to $[1]$$[1]$ in each category is:

• $\mathrm{Top}$$\mathrm{Top}$: the one-point discrete space $\{\ast \}$$\{*\}$

• $\mathrm{Mon}$$\mathrm{Mon}$: $(\mathbb{N},+)$$(\mathbb{N}, +)$

• $\mathrm{Grp}$$\mathrm{Grp}$: $(\mathbb{Z},+)$$(\mathbb{Z}, +)$

• $\mathrm{Ring}$$\mathrm{Ring}$: $\mathbb{Z}[X]$$\mathbb{Z}[X]$, or more generally, $R$$R$-alg, $R[X]$$R[X]$

• $R\text{-}\mathrm{Mod}$$R\text{-}\mathrm{Mod}$: the free $R$$R$-module $R$$R$

---

Lemma. Let $F$$F$ be a faithful functor. Then if $F(f)$$F(f)$ is a monomorphism, it follows that $f$$f$ is a monomorphism.

Proof.
Suppose $F(f)$$F(f)$ is a monomorphism. Then:

$$f\circ g=f\circ h⟹F(f)\circ F(g)=F(f)\circ F(h)⟹F(g)=F(h)⟹g=h◻$$

---

Lemma. Let $\mathcal{C}$$\mathcal{C}$ be a locally small category. Then if $f:X\to Y$$f: X \to Y$ is a monomorphism, the induced map:

$$f_{\ast }:{\mathrm{Hom}}_{\mathcal{C}}(A,X)\to {\mathrm{Hom}}_{\mathcal{C}}(A,Y)$$

is injective for all $A$$A$.

Proof.

If $f_{\ast }(g)=f_{\ast }(h)$$f_*(g) = f_*(h)$, then $g=h$$g = h$ by the definition of a monomorphism. $◻$$\square$

---

So, $f$$f$ is a monomorphism in $\mathcal{C}$$\mathcal{C}$ if and only if ${\mathrm{Hom}}_{\mathcal{C}}(F([1]),f)$$\mathrm{Hom}_{\mathcal{C}}(F([1]), f)$, i.e., $f_{\ast }$$f_*$, is injective. This follows from the fact that the forgetful functor is both faithful and representable, so $f$$f$ is mono iff $f$$f$ is injective on underlying set. $◻$$\square$

Remark. Actually you can prove $f$$f$ is mono iff $f$$f$ is injective in $\mathrm{Set}$$\mathrm{Set}$ as well, both free and forgetful functor are identity.

Remark. Remeber the category $\mathcal{N}$$\mathcal N$?

The forgetful functor is representable by ${\mathrm{Hom}}_{{\mathbb{N}}^{+}}(1,-)$$\mathrm{Hom}_{\mathbb N^+}(1,-)$! Hence in this category, mono iff injective as well.