Let \(\mathcal{S}\) be the category of short exact sequences:

\(\begin{equation} 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 \end{equation}\)

in \(\mathcal{A}\). Show that \(\delta_i\) is a natural transformation from the functor sending the above SES to \(T_{i}(C)\) to the functor sending the SES to \(T_{i-1}(A)\).

Given a short exact sequence as above and \(\delta_i\), a \(\delta\)-functor. Observe that from the second part of the definition of \(\delta\)-functors, the diagram of the identity morphism of the SES above must commute. Thus, satisfying the definition of a natural transformation from the functor sending our SES to \(T_i(C)\) to the functor sending the SES to \(T_{i-1}(C)\).

(is this the right start to the idea?)

Show a chain complex \(P\) is a projective object in \(Ch\) if and only if it is a split exact complex of projectives.

Solving the only if part first:

Following the hint where we consider \(\begin{equation} 0 \ldots \to P_1 \cong P_0 \to 0 \to \ldots \end{equation}\).

Since \(P\) is split exact, we can decompose \(P_n\) into \(P_n = Z_n \oplus B_n\). Where \(Z_n = ker(d_n), B_n = P_n \ Z_n\) Thus, we want to decompose all of \(P\) in such a way. Define a complex \(H._{n}\) to be \(\ldots \to 0 \to B_{n} \to Z_{n-1} \to 0 \to \ldots \) for each \(P_n\). Taking a direct sum of all the \(H._{n}\) recovers \(P\).

Now, if we had maps \(f: P \to Y\) and \(\pi : X \to Y\) where \(\pi\) is a surjection, then the map \(f\) can be reinterpreted as a decomposed map, \(f= \sum f_i\) acting on each \(H._{n}\). Projectivity at each \(P_n\) yields that the \(f_i\) must be projective on \(f_i\) level, giving us maps \(g_i\) such that \(\pi \circ g_i = f_i\). Define \(g:=\sum g_i\). We see then, that \(pi \circ g= f\). Hence, \(P\) is a projective object.

For the forwards direction:

Let \(P\) be a projective object. Per the hint, we consider the surjection from \(\textrm{cone}(id_p)\) to \(P[-1]\). We expand this into a short exact sequence as follows:

\(\begin{equation} 0 \to P \to \textrm{cone}(id) \to P[-1] \to 0 \end{equation}\)

where the first map sends \(p \in P\) to \((0,p)\) and the second map, \(\pi\), sends \((b,p) \to -b\).

If \(P\) is projective, it is clear that \(P[-1]\) is too. Thus, with the surjection from \(\textrm{cone}(id) \to P[-1]\), and the identity map on \(P[-1]\) to itself, we apply projectivity of \(P[-1]\) and obtain a chain map \(h:P[-1] \to \textrm{cone}(id)\) with \(h(p) = (-p, f(p) )\) such that \(id = \pi \circ h\)

Now, notice that \(h\) being a chain map implies that \(dh=hd\). Meaning,

\(\begin{align} d(h(p)) = d(-p,f(p) ) &= (-d(-p), d(f(p)) +p) \\ &= (-d(p), f(d(p)) ) \\ &= (d(p), f(-p)) \\ &= h(-p) \\ &= h(d(p)) \end{align}\)

Inside these equalities, we see that \(d(f(p)) - p = f(d(p))\). Making \(f\) a chain contraction of the identity map on \(P\). Thus, \(P\) is a split exact complex.

To see why each \(P_n\) is projective, notice that by split exactness of \(P\), we get decomposition \(P_n = Im(sd) \oplus coker(sd)\). Notice how \(d\) kills the first set and maps the second to \(P_{n-1}\).

We can then apply the projectivity of \(P\) to maps \(f: P \to Y\) and \(\pi : X \to Y\) where \(\pi\) is a surjection. Where \(X\) and \(Y\) are the chain complex with zeroes everywhere except for a module on the same \(n\)th coordinate where \(coker(sd)\) lies. Projectivity of \(P\) forces \(P_n\) to be a projective object as well.

Let \(R = \mathbb{Z} / m\). Use Baer's criterion to show that \(R\) is an injective \(R\)-module. Then show that \(\mathbb{Z}/d\) is not an injective R-module when \(d|m\) and some prime \(p\) divides both \(d\) and \(m/d\).

Let \(R\) as above.