I was reading Mark Hovey's Model Categories and I am confused about the following proof for left homotopy being an equivalence relation
Firstly, how does $B$ being cofibrant imply that $t$ is a weak equivalence? I was thinking maybe we could show it is the retract of a weak equivalence, or use the "2-of-3 axiom".
Secondly, are the maps $j_0$ and $j_1$ defined correctly, or are the indicies on the $i$'s switched?
Since $s$ is always a weak equivalence, the only interesting statement here is that the map $B^\prime \to C$ (and, by the same argument, also $B^{\prime\prime} \to C$) is a trivial cofibration. This is because the morphism $t$, which is induced by the universal property for $C$, ensures that the diagram $$\require{AMScd} \begin{CD} B^\prime @>>> C \\ @VsVV @VtVV\\ B @>1_B>> B \end{CD}$$ commutes so that $t$ is a weak equivalence by 2-out-of-3.
For the interesting statement, we observe that because $B$ is cofibrant and cofibrations are stable under pushout, both the coproduct inclusions $$\require{AMScd} \begin{CD} 0 @>>> B\\ @VVV @VVV\\ B @>>> B\coprod B \end{CD}$$ are cofibrations. The morphism $i_0^\prime \colon B \to B^{\prime\prime}$ is the composition of the cofibration $B \coprod B \to B^{\prime\prime}$ with one of these inclusions, hence is itself a cofibration. Moreover, we observe that $i_0^\prime$ is a weak equivalence because the morphism $s^\prime \colon B^{\prime\prime} \to B$ is a weak equivalence and $s \circ i_0^\prime = 1_B.$
Now apply stability of trivial cofibrations under pushout to the commutative diagram in question to see that $B^\prime \to C$ is a trivial cofibration.