Free Modules, Exact Sequences, Projective and Injective Modules

Define $f((m_i)):={\sum }_{i\in \mathrm{supp}(m)}{f}_i(m_i)$; the sum is finite by [F1], and padding it by zero terms shows it is independent of the chosen finite set containing the support.

Addition and scalar multiplication may be checked on the finite union of the relevant supports, so [F2] gives $f(m+m^{\mathrm{\prime }})=f(m)+f(m^{\mathrm{\prime }})$ and $f(rm)=rf(m)$.

For $x\in M_i$, step 1.1 gives $f({\mathrm{ȷ}}_i(x))=f_i(x)$, so $f\circ {\mathrm{ȷ}}_i=f_i$.

Every $m=(m_i)$ equals the finite sum ${\sum }_{i\in \mathrm{supp}(m)}{\mathrm{ȷ}}_i(m_i)$. Hence any homomorphism $g$ satisfying $g\circ {\mathrm{ȷ}}_i=f_i$ has $g(m)={\sum }_i{f}_i(m_i)=f(m)$ and therefore equals $f$.

If $I=\mathrm{\varnothing }$, the direct sum is $0$ by [F1], the formula is the empty sum, and the construction and uniqueness still apply. Thus the universal property holds for every index set.

For each $x\in X$, define the homomorphism $f_x:R\to M$ by $f_x(r)=ru(x)$.

By [L1], the family $(f_x)$ determines a unique homomorphism $\bar{u}:R^{(X)}\to M$ with $\bar{u}\circ {\mathrm{ȷ}}_x=f_x$.

Since $e_x={\mathrm{ȷ}}_x(1_R)$, one has $\bar{u}(e_x)=f_x(1_R)=u(x)$, and additivity gives the displayed finite-sum formula.

Any homomorphism agreeing with $u$ on every $e_x$ agrees with $\bar{u}$ on every finite linear combination, hence on all of $R^{(X)}$.

When $X=\mathrm{\varnothing }$, [F1] gives $R^{(X)}=0$ and the unique map $0\to M$, so no nonempty choice is hidden. The construction and uniqueness prove the universal property.

Apply [L1] to the identity set map on the underlying set of $M$; this defines ${\epsilon }_M:R^{(M)}\to M$ with ${\epsilon }_M(e_m)=m$.

Every $m\in M$ equals ${\epsilon }_M(e_m)$, so ${\epsilon }_M$ is surjective, including when $M=0$.

Define $\varphi :R^{(M)}\mathrm{/}\ker {\epsilon }_M\to M$ by $\varphi (x+\ker {\epsilon }_M)={\epsilon }_M(x)$. Equality of cosets makes this well defined, and [F1] makes it a homomorphism.

The map $\varphi$ is surjective by step 2.1 and injective because $\varphi (x+\ker {\epsilon }_M)=0$ exactly when $x\in \ker {\epsilon }_M$. Hence it is an isomorphism.

Thus $M$ is canonically a quotient of a free module.

Record the images of the standard basis vectors from [F4] as the columns of rectangular matrices $A$ and $B$. The coordinate formula and [F2] turn the two inverse composites into $AB=I_n$ and $BA=I_m$.

Suppose first that $n>m$. Each of the $n$ columns of $AB$ is an $R$-linear combination of the $m$ columns of $A$.

Expanding $\det (AB)$ by multilinearity in all $n$ columns, each term chooses one of the $m$ columns of $A$ in each of $n$ positions. Since $n>m$, some chosen column repeats, so every term is zero by [L1]. Thus $\det (AB)=0$.

But $AB=I_n$, so normalization gives $\det (AB)=1_R\mathrm{\ne }0$, contradicting step 3.1. Hence $n\le m$.

Interchanging $A,B$ and $m,n$ gives $m\le n$. Therefore $m=n$, proving [F1]. The cases $m=0$ or $n=0$ are included: a strict inequality makes the other positive and the same determinant argument applies.

The zero map $0\to A$ has image $\{0\}$, so by [F1] the sequence $0\to A\stackrel{f}{\to }B$ is exact at $A$ exactly when $\ker f=\{0\}$, which is equivalent to injectivity by [L1].

The zero map $B\to 0$ has kernel $B$, so by [F1] the sequence $A\stackrel{f}{\to }B\to 0$ is exact at $B$ exactly when $\mathrm{im}f=B$, which is equivalent to surjectivity by [F2].

A four-term sequence $0\to A\stackrel{i}{\to }B\stackrel{p}{\to }C\to 0$ is exact at $A,B,C$ exactly when the endpoint conditions of steps 1.1 and 1.2 hold and $\mathrm{im}i=\ker p$ holds at $B$.

This proves both endpoint equivalences and both directions of the short-exact characterization.

Suppose $s$ is a section and define $\mathrm{\Phi }:A\oplus C\to B$ by $\mathrm{\Phi }(a,c)=i(a)+s(c)$; [L2] makes this a homomorphism.

Suppose assertion 3 holds. Define $r:B\to A$ as the first coordinate of ${\mathrm{\Phi }}^{-1}$; then $\mathrm{\Phi }(a,0)=i(a)$ gives $r(i(a))=a$, so $r$ is a retraction.

Suppose instead that $r$ is a retraction. For each $c\in C$, choose any $b$ with $p(b)=c$ using surjectivity and put $k=b-i(r(b))$. Then $r(k)=0$ and $p(k)=c$.

For $b\in B$, the element $b-s(p(b))$ lies in $\ker p=\mathrm{im}i$, so by injectivity of $i$ there is a unique $a\in A$ with $i(a)=b-s(p(b))$; hence $b=\mathrm{\Phi }(a,p(b))$ and $\mathrm{\Phi }$ is surjective.

If $\mathrm{\Phi }(a,c)=0$, applying $p$ gives $c=0$, and then injectivity of $i$ gives $a=0$; thus $\mathrm{\Phi }$ is injective and satisfies the compatibility conditions in assertion 3.

The element $k$ of step 1.3 is unique in $\ker r$ with image $c$: if $k^{\mathrm{\prime }}\in \ker r$ and $p(k^{\mathrm{\prime }})=c$, then $k-k^{\mathrm{\prime }}\in \ker p=\mathrm{im}i$, say $k-k^{\mathrm{\prime }}=i(a)$, and applying $r$ gives $a=0$. Therefore the rule $s(c)=k$ is independent of the temporary lift $b$, is linear by uniqueness, and satisfies $p\circ s={\mathrm{id}}_C$.

Steps 1.1, 2.1, and 2.2 prove $1\Rightarrow 3$, step 1.2 proves $3\Rightarrow 2$, and steps 1.3 and 2.3 prove $2\Rightarrow 1$. The formula for $\mathrm{\Phi }$ also yields the internal direct sum $B=i(A)\oplus s(C)$.

If $u_{\ast }f=0$, then $u(f(x))=0$ for every $x$, and injectivity of $u$ gives $f=0$; hence $u_{\ast }$ is injective.

Composability gives $v_{\ast }{u}_{\ast }=0$, so $\mathrm{im}{u}_{\ast }\le \ker v_{\ast }$.

If $g:X\to B$ satisfies $v_{\ast }g=0$, then $g(x)\in \ker v=\mathrm{im}u$ for every $x$. Injectivity of $u$ gives a unique $f(x)\in A$ with $u(f(x))=g(x)$; uniqueness makes $f$ linear, so $g=u_{\ast }f$.

If $p^{\ast }h=0$, surjectivity of $p$ gives $h=0$, so $p^{\ast }$ is injective; and $u^{\ast }{p}^{\ast }=0$ because $p\circ u=0$.

If $g:B\to X$ satisfies $u^{\ast }g=g\circ u=0$, then $g$ vanishes on $\mathrm{im}u=\ker p$. By [L1] it factors uniquely through $B\mathrm{/}\ker p$, and since $p$ is surjective the rule $\bar{g}(p(b))=g(b)$ gives the corresponding homomorphism $\bar{g}:C\to X$ with $g=\bar{g}\circ p=p^{\ast }(\bar{g})$.

Steps 1.1 to 1.3 prove the covariant sequence exact, and steps 1.4 and 1.5 prove the contravariant sequence exact.

Assume $a$ is surjective and $b,d$ are injective, and let $z\in C$ satisfy $c(z)=0$. Then [C3] gives $d(w(z))=w^{\mathrm{\prime }}(c(z))=0$, so injectivity of $d$ gives $w(z)=0$.

Assume $b,d$ are surjective and $e$ is injective, and let $z^{\mathrm{\prime }}\in C^{\mathrm{\prime }}$. By surjectivity of $d$, choose $q\in D$ with $d(q)=w^{\mathrm{\prime }}(z^{\mathrm{\prime }})$.

Exactness gives $y\in B$ with $v(y)=z$. By [C2], $v^{\mathrm{\prime }}(b(y))=c(v(y))=0$, so exactness gives $y^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$ with $u^{\mathrm{\prime }}(y^{\mathrm{\prime }})=b(y)$.

By [C4], $e(x(q))=x^{\mathrm{\prime }}(d(q))=x^{\mathrm{\prime }}{w}^{\mathrm{\prime }}(z^{\mathrm{\prime }})=0$; injectivity of $e$ gives $x(q)=0$. Exactness gives $z\in C$ with $w(z)=q$.

Surjectivity of $a$ gives $y^{\mathrm{\prime }}=a(t)$ for some $t\in A$. Then [C1] gives $b(y-u(t))=0$; injectivity of $b$ gives $y=u(t)$, and exactness gives $z=v(y)=v(u(t))=0$. Thus $c$ is injective.

Now $w^{\mathrm{\prime }}(c(z)-z^{\mathrm{\prime }})=d(w(z))-w^{\mathrm{\prime }}(z^{\mathrm{\prime }})=0$ by [C3], so exactness gives $y^{\mathrm{\prime }}\in B^{\mathrm{\prime }}$ with $v^{\mathrm{\prime }}(y^{\mathrm{\prime }})=c(z)-z^{\mathrm{\prime }}$. Surjectivity of $b$ gives $y^{\mathrm{\prime }}=b(y)$; then [C2] yields $c(z-v(y))=z^{\mathrm{\prime }}$. Thus $c$ is surjective.

Steps 1.1, 2.1, and 3.1 prove the injective Four Lemma, while steps 1.2, 2.2, and 3.2 prove the surjective Four Lemma.

Under the first set of hypotheses, the injective Four Lemma [L1] applied to the diagram [C1] to [C4] gives that $c$ is injective.

Under the second set of hypotheses, the surjective Four Lemma [L1] applied to the same diagram gives that $c$ is surjective.

If $a,b,d,e$ are isomorphisms, then $a,b,d$ meet the first hypotheses and $b,d,e$ meet the second; steps 1.1 and 1.2 make $c$ both injective and surjective, hence an isomorphism.

The restrictions $\ker a\to \ker b$ and $\ker b\to \ker c$ are induced by $i$ and $p$ using [C1] and [C2]. The formulas $[x^{\mathrm{\prime }}]\mapsto [i^{\mathrm{\prime }}(x^{\mathrm{\prime }})]$ and $[y^{\mathrm{\prime }}]\mapsto [p^{\mathrm{\prime }}(y^{\mathrm{\prime }})]$ define maps $\mathrm{coker}a\to \mathrm{coker}b$ and $\mathrm{coker}b\to \mathrm{coker}c$: [C1] and [C2] make the relevant images vanish in the target quotients, so [L1] applies.

For $z\in \ker c$, choose $y\in B$ with $p(y)=z$. Then [C2] gives $p^{\mathrm{\prime }}(b(y))=c(z)=0$, so [F1] gives a unique $x^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$ with $i^{\mathrm{\prime }}(x^{\mathrm{\prime }})=b(y)$. Define $\delta (z):=[x^{\mathrm{\prime }}]\in \mathrm{coker}a$.

If $\tilde{y}$ is another lift of $z$, then $\tilde{y}-y=i(x)$ for some $x\in A$ by [F1]. If $i^{\mathrm{\prime }}({\tilde{x}}^{\mathrm{\prime }})=b(\tilde{y})$, then [C1] and injectivity of $i^{\mathrm{\prime }}$ give ${\tilde{x}}^{\mathrm{\prime }}-x^{\mathrm{\prime }}=a(x)$, so $[{\tilde{x}}^{\mathrm{\prime }}]=[x^{\mathrm{\prime }}]$ in $\mathrm{coker}a$. Thus $\delta$ is well defined.

Exactness at $\ker a$ holds because its map is the restriction of the injective map $i$. At $\ker b$, the composite induced by $p\circ i$ is zero; if $y\in \ker b$ maps to zero in $\ker c$, then $p(y)=0$, so $y=i(x)$ by [F1], and [C1] with injectivity of $i^{\mathrm{\prime }}$ gives $a(x)=0$, hence $x\in \ker a$.

If $y\in \ker b$, the construction of step 1.2 applied to $z=p(y)$ has $x^{\mathrm{\prime }}=0$, so $\delta (z)=0$. Conversely, if $z\in \ker c$ has $\delta (z)=0$, choose $y,x^{\mathrm{\prime }}$ as in step 1.2; then $x^{\mathrm{\prime }}=a(x)$ for some $x$, so [C1] gives $b(y-i(x))=0$ and $p(y-i(x))=z$. Thus exactness holds at $\ker c$.

The map $\mathrm{coker}a\to \mathrm{coker}b$ kills $\delta (z)$ because $i^{\mathrm{\prime }}(x^{\mathrm{\prime }})=b(y)$. Conversely, if $[x^{\mathrm{\prime }}]$ maps to zero, write $i^{\mathrm{\prime }}(x^{\mathrm{\prime }})=b(y)$; then [C2] gives $c(p(y))=0$, and the construction with lift $y$ gives $\delta (p(y))=[x^{\mathrm{\prime }}]$. Thus exactness holds at $\mathrm{coker}a$.

The next composite is zero because $p^{\mathrm{\prime }}{i}^{\mathrm{\prime }}=0$. If $[y^{\mathrm{\prime }}]\in \mathrm{coker}b$ maps to zero in $\mathrm{coker}c$, write $p^{\mathrm{\prime }}(y^{\mathrm{\prime }})=c(z)$, choose $y\in B$ with $p(y)=z$, and use [C2] to obtain $y^{\mathrm{\prime }}-b(y)\in \ker p^{\mathrm{\prime }}=\mathrm{im}{i}^{\mathrm{\prime }}$. Hence $[y^{\mathrm{\prime }}]$ comes from $\mathrm{coker}a$, proving exactness at $\mathrm{coker}b$.

The map $\mathrm{coker}b\to \mathrm{coker}c$ is surjective: for a class $[z^{\mathrm{\prime }}]$, choose $y^{\mathrm{\prime }}\in B^{\mathrm{\prime }}$ with $p^{\mathrm{\prime }}(y^{\mathrm{\prime }})=z^{\mathrm{\prime }}$ using [F1], and $[y^{\mathrm{\prime }}]$ maps to $[z^{\mathrm{\prime }}]$.

For $z_1,z_2\in \ker c$ and $r\in R$, the choices $y_1+y_2$ and $r{y}_1$ in step 1.2 lead to $x_1^{\mathrm{\prime }}+x_2^{\mathrm{\prime }}$ and $r{x}_1^{\mathrm{\prime }}$; uniqueness through the injective map $i^{\mathrm{\prime }}$ then gives $\delta (z_1+z_2)=\delta (z_1)+\delta (z_2)$ and $\delta (r{z}_1)=r\delta (z_1)$.

Steps 2.2 through 2.6 establish exactness at every displayed term. Steps 1.2, 2.1, and 3.1 construct a well-defined linear connecting homomorphism.

For each $x\in X$, the fiber $q^{-1}(f(e_x))$ is nonempty because $q$ is surjective.

Under AC, [F2] chooses $y_x\in q^{-1}(f(e_x))$ for every $x\in X$. If $X$ is finite with a given finite enumeration, [L2] makes this choice in ZF; if $X=\mathrm{\varnothing }$, there are no choices.

By [L1], the assignment $e_x\mapsto y_x$ extends uniquely to a homomorphism $\tilde{f}:F\to E$.