I am reading an example related to discrete-time Markov chains which I do not really understand rigorously.
Suppose that $\{ X_n \}_{n \in \mathbb{N} }$ is a time-homogeneous discrete-time Markov chain, with state space $S$. As usual, we define the first-passage time of state $s$ to be $$ T_s = \inf \{ n \geq 1: X_n =s \}.$$ We fix two distinct states $s_1, s_2 \in S$. We then define $b_i := \mathbb{P} ( T_{s_1} < T_{s_2} \, \big| \, X_0 =i ).$ It is claimed that
\begin{eqnarray} b_i &= & \sum_{j \in S} \mathbb{P} (T_{s_1} < T_{s_2} , X_1 =j \, \big| \, X_0 =i ) \\ &= & \sum_{j \in S} \mathbb{P} (T_{s_1} < T_{s_2} \, \big| \, X_1 =j, X_0 =i ) \, \mathbb{P} ( X_1 = j \, \big| \, X_0 =i) \\ &= & \sum_{j \in S} b_j \mathbb{P} ( X_1 = j \, \big| \, X_0 =i) . \end{eqnarray}
I don't really understand why $\mathbb{P} (T_{s_1} < T_{s_2} \, \big| \, X_1 =j, X_0 =i ) = b_j.$ We can apply Markov's property and obtain that $$\mathbb{P} (T_{s_1} < T_{s_2} \, \big| \, X_1 =j, X_0 =i ) = \mathbb{P} (T_{s_1} < T_{s_2} \, \big| \, X_1=j).$$ But that still doesn't mean that we can change the time point from $1$ to $0$. Is it time-homogeneity that was used?
You are right to be skeptical as there are some conditions missing.
For example, $\mathbb{P}(T_{s_1}<T_{s_2}\mid X_1=s_1)=1$, but there is no guarantee that $b_{s_1}=1$.
The equation $b_i=\sum_{j\in S}b_j\, p_{ij}$ will be true if the initial state $i$ is so "far away" from $s_1$ that $X_1=s_1$ is impossible.