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In the direct sum of modules $M=N\oplus P$, every $m\in M$ has a unique expression $m=n+p$, with $n\in N$ and $p\in P$. DefineThe uniqueness makes this well-defined. For $m=n+p$, $m^{\prime }=n^{\prime }+p^{\prime }$ and $r\in R$,so $\pi$ is a module homomorphism. Finally, $\pi (n+p)=0$ exactly when $p=0$, equivalently $m\in N$. Therefore $\boxed{\ker \pi =N}$. Its image is $P$, and ${\pi }^2=\pi$, so it is the projection onto that direct summand.

Under the temporary assumption that $E$ is a basis of a module for $B$, part (a)(v) applied to the countably generated submodule $F$ gives a countable set $E^{\prime }\subseteq E$ such that $F\subseteq C=⟨E^{\prime }⟩$. By part (a)(iv),Use part (a)(iii) to project onto the second direct summand. This gives a module homomorphism $\pi :B\to B$ withIt remains to check the required size of this kernel. Enumerate $E^{\prime }$ as $b_1,b_2,\dots$, with the finite case treated the same way. Every element of $C$ is a finite sum of these elements with integer coefficients. For each $m$, the set $⟨b_1,\dots ,b_m⟩$ is the image of the countable set ${\mathbb{Z}}^m$, so it is countable. Taking their countable union provesThis is the countability of countably generated abelian groups argument; it uses the countability of the scalar ring $\mathbb{Z}$.

A projective module is a direct summand of a free module. Free modules are flat because tensoring with a free module produces a direct sum of copies of the original module and map. If $F=P\oplus Q$ is flat, tensoring an injection with $F$ is the direct sum of its tensor maps with $P$ and $Q$, so each component is injective. Applying this to a free module containing $P$ proves that $P$ is flat.

In an abelian category with enough projectives, a projective object has a resolution concentrated in degree zero, so every first left derived functor vanishes on it. Conversely choose $0\to K\to P\to A\to 0$ with $P$ projective and use the additive functor $F=\mathcal{A}(-,K)$ with target the opposite of the category of abelian groups. This functor is right exact in that target. If $L_{1}F(A)=0$, the derived exact sequence makes $F(K)\to F(P)$ monic, meaning that $\mathrm{Hom}(P,K)\to \mathrm{Hom}(K,K)$ is surjective in abelian groups. Extending $1_K$ gives a retraction of $K\to P$, so the sequence splits and $A$ is a direct summand of the projective $P$.

The first integral singular homology of the Hawaiian earring has a direct summand isomorphic to $\mathbb{Q}$. This consequence of its structural homology theorem suffices, through the universal coefficient theorem for cohomology and nonzero Ext of the rationals with integer coefficients, to show that its integral $H^2$ is nonzero. The precise structural theorem is Theorem 3.1 of www.sas.rochester.edu/mth/sites/doug-ravenel/otherpapers/eda-kawamura2.pdf.

For an indecomposable module $M$ with vertex of an indecomposable module $D$, a source is an indecomposable $RD$-module $S$ which is a direct summand of ${\mathrm{Res}}_D^{G}M$ and for which $M$ is a direct summand of ${\mathrm{Ind}}_D^{G}S$. It has vertex $D$ as a $D$-module. Sources at a fixed vertex are conjugate by its normalizer.

Over a modular field or a complete discrete valuation coefficient ring, an indecomposable module has trivial source when a source is the rank-one trivial module of its vertex. These are precisely the indecomposable direct summands of permutation modules. A general trivial source module is a direct sum of these indecomposables.

Integer row vectors forming a square unimodular matrix are a basis of the corresponding free abelian group: the adjugate formula makes the inverse integral. The subgroup spanned by any subset of that basis is therefore a direct summand, complemented by the remaining basis vectors. In contrast, a proper finite-index subgroup of a free abelian group cannot be a direct summand: its nonzero finite quotient would have to embed as a subgroup of a torsion-free group.