= Solution
{c}
We prove $(i)\Rightarrow(iii)$, the main step in <Ext-connected components determine blocks>. Suppose a <block of a finite-dimensional algebra> $B=Re$ had more than one component of the graph of nonsplit extensions between its <simple modules>. Partition its simple types into one component and the union of the others. There are no cross-extensions, so part (a) decomposes the left regular $B$-module as $B=U_1\oplus U_2$. Both summands are nonzero: every simple $B$-module is a quotient of the regular module and therefore occurs among its composition factors.
The maximality in (a) makes these summands canonical. Every right multiplication is a left $B$-module endomorphism, so it preserves them. Thus they are two-sided ideals, and their projection $p:B\to U_1$ commutes with both left and right multiplication. Putting $f=p(e)$ gives
$$
p(x)=xf=fx\quad(x\in B),\qquad f^2=f.
$$
Since both summands are nonzero, $f\ne0,e$. This contradicts the primitivity of the central block idempotent $e$. There is therefore just one Ext component within each block, proving that simples in the same block satisfy $\boxed{S\sim T}$.
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