[proofplan] The transition function $t_{ij}$ is the fibre component of the coordinate change $\phi_i \circ \phi_j^{-1}$ over the overlap $U_i \cap U_j$; this defining relation packages the base-preserving property of the bundle trivialisations. The identity and inverse laws follow from the identity map $\phi_i \circ \phi_i^{-1}$ and from inverting the coordinate change $\phi_i \circ \phi_j^{-1}$. The triple-overlap identity follows by factoring the coordinate change from the $k$-trivialisation to the $i$-trivialisation through the $j$-trivialisation and comparing the fibre components. [/proofplan] [step:Declare the trivialising cover and define the transition functions] Let $I$ denote the index set of the chosen trivializing cover $(U_i)_{i\in I}$ of $B$, and let $F$ denote the typical fibre of the fibre bundle. For each $i \in I$, let \begin{align*} \phi_i: \pi^{-1}(U_i) &\to U_i \times F \end{align*} denote the chosen local trivialisation. For $i,j \in I$ with $U_i \cap U_j \neq \varnothing$, the transition function \begin{align*} t_{ij}: U_i \cap U_j &\to \operatorname{Homeo}(F) \end{align*} is defined by the identity \begin{align*} \phi_i \circ \phi_j^{-1}(b,x)=(b,t_{ij}(b)(x)) \end{align*} for every $(b,x) \in (U_i \cap U_j)\times F$. [/step] [step:Compare a trivialisation with itself to obtain the identity transition function] Fix $i \in I$ and $b \in U_i$. By definition of $t_{ii}: U_i \to \operatorname{Homeo}(F)$, for every $x \in F$, \begin{align*} \phi_i \circ \phi_i^{-1}(b,x) = (b, t_{ii}(b)(x)). \end{align*} Since $\phi_i \circ \phi_i^{-1} = \operatorname{id}_{U_i \times F}$, the same left-hand side is $(b,x)$. Therefore $(b,t_{ii}(b)(x))=(b,x)$ for every $x \in F$, hence $t_{ii}(b)(x)=x$ for every $x \in F$. Thus \begin{align*} t_{ii}(b)=\operatorname{id}_F. \end{align*} [/step] [step:Invert a coordinate change to obtain the inverse transition function] Fix $i,j \in I$ with $U_i \cap U_j \neq \varnothing$, and fix $b \in U_i \cap U_j$. For every $x \in F$, the definition of $t_{ij}$ gives \begin{align*} \phi_i \circ \phi_j^{-1}(b,x) = (b,t_{ij}(b)(x)). \end{align*} The inverse homeomorphism of $\phi_i \circ \phi_j^{-1}$ is $\phi_j \circ \phi_i^{-1}$. Therefore, applying $\phi_j \circ \phi_i^{-1}$ to both sides gives \begin{align*} (b,x)=\phi_j \circ \phi_i^{-1}(b,t_{ij}(b)(x)). \end{align*} By the definition of $t_{ji}: U_i \cap U_j \to \operatorname{Homeo}(F)$, this becomes \begin{align*} (b,x)=(b,t_{ji}(b)(t_{ij}(b)(x))). \end{align*} Thus $t_{ji}(b)\circ t_{ij}(b)=\operatorname{id}_F$. Interchanging $i$ and $j$ gives $t_{ij}(b)\circ t_{ji}(b)=\operatorname{id}_F$. Hence $t_{ij}(b)$ and $t_{ji}(b)$ are inverse homeomorphisms of $F$, so \begin{align*} t_{ij}(b)=t_{ji}(b)^{-1}. \end{align*} [/step] [step:Factor a coordinate change through a third trivialisation on triple overlaps] Fix $i,j,k \in I$ with $U_i \cap U_j \cap U_k \neq \varnothing$, and fix $b \in U_i \cap U_j \cap U_k$. All coordinate changes in this step are restricted to $(U_i \cap U_j \cap U_k)\times F$, where the maps $\phi_i \circ \phi_j^{-1}$, $\phi_j \circ \phi_k^{-1}$, and $\phi_i \circ \phi_k^{-1}$ are well-defined. For every $x \in F$, the coordinate change from the $k$-trivialisation to the $i$-trivialisation factors as \begin{align*} \phi_i \circ \phi_k^{-1} = (\phi_i \circ \phi_j^{-1}) \circ (\phi_j \circ \phi_k^{-1}) \end{align*} on $\{b\}\times F$. Applying the right-hand side to $(b,x)$ and using the definition of $t_{jk}$ gives \begin{align*} (\phi_i \circ \phi_j^{-1}) \circ (\phi_j \circ \phi_k^{-1})(b,x)=(\phi_i \circ \phi_j^{-1})(b,t_{jk}(b)(x)). \end{align*} Using the definition of $t_{ij}$ on this last expression gives \begin{align*} (\phi_i \circ \phi_j^{-1})(b,t_{jk}(b)(x))=(b,t_{ij}(b)(t_{jk}(b)(x))). \end{align*} Applying the direct coordinate change gives \begin{align*} \phi_i \circ \phi_k^{-1}(b,x)=(b,t_{ik}(b)(x)). \end{align*} Comparing second components, we obtain \begin{align*} t_{ij}(b)(t_{jk}(b)(x))=t_{ik}(b)(x) \end{align*} for every $x \in F$. Therefore \begin{align*} t_{ij}(b)\circ t_{jk}(b)=t_{ik}(b). \end{align*} [guided] The point of the triple-overlap identity is that there are two ways to express the same coordinate change from the $k$-chart to the $i$-chart. Recall that for each index $r \in I$, the map $\phi_r: \pi^{-1}(U_r) \to U_r \times F$ is a chosen local trivialisation, and for $r,s \in I$ the transition function $t_{rs}: U_r \cap U_s \to \operatorname{Homeo}(F)$ is defined by $\phi_r \circ \phi_s^{-1}(b,x)=(b,t_{rs}(b)(x))$ on $(U_r\cap U_s)\times F$. Fix indices $i,j,k \in I$ with $U_i \cap U_j \cap U_k \neq \varnothing$, and fix a base point $b \in U_i \cap U_j \cap U_k$. We compare maps on the fibre $\{b\}\times F$, with all coordinate changes restricted to $(U_i \cap U_j \cap U_k)\times F$ so that the compositions are well-defined. For each $x \in F$, the direct coordinate change from the $k$-trivialisation to the $i$-trivialisation is \begin{align*} \phi_i \circ \phi_k^{-1}(b,x)=(b,t_{ik}(b)(x)). \end{align*} Now perform the same coordinate change in two stages: first go from the $k$-trivialisation to the $j$-trivialisation, then from the $j$-trivialisation to the $i$-trivialisation. Since composition of maps is associative, this two-stage coordinate change is \begin{align*} (\phi_i \circ \phi_j^{-1}) \circ (\phi_j \circ \phi_k^{-1}) = \phi_i \circ (\phi_j^{-1}\circ \phi_j) \circ \phi_k^{-1}. \end{align*} Since $\phi_j^{-1}\circ \phi_j$ is the identity on its domain, this simplifies to \begin{align*} (\phi_i \circ \phi_j^{-1}) \circ (\phi_j \circ \phi_k^{-1})=\phi_i \circ \phi_k^{-1}. \end{align*} Evaluating the two-stage expression at $(b,x)$ and first using the definition of $t_{jk}$ gives \begin{align*} (\phi_i \circ \phi_j^{-1}) \circ (\phi_j \circ \phi_k^{-1})(b,x)=(\phi_i \circ \phi_j^{-1})(b,t_{jk}(b)(x)). \end{align*} Then using the definition of $t_{ij}$ gives \begin{align*} (\phi_i \circ \phi_j^{-1})(b,t_{jk}(b)(x))=(b,t_{ij}(b)(t_{jk}(b)(x))). \end{align*} The direct and two-stage coordinate changes are the same map, so their values at $(b,x)$ agree: \begin{align*} (b,t_{ik}(b)(x))=(b,t_{ij}(b)(t_{jk}(b)(x))). \end{align*} Equality in the product $U_i \times F$ implies equality of second components, hence \begin{align*} t_{ik}(b)(x)=t_{ij}(b)(t_{jk}(b)(x)). \end{align*} Because this holds for every $x \in F$, the two maps $F \to F$ are equal: \begin{align*} t_{ij}(b)\circ t_{jk}(b)=t_{ik}(b). \end{align*} [/guided] [/step]