In the category of simplicial set $S$, we have a bijection
$ev_*:\hom_S(K,\operatorname{Hom}(X,Y))\to \hom_S(X\times K , Y)$
I wonder how to identify commutative diagrams using this bijection. Specifically how to identify the two diagrams in the pictures.
Given $i:K\to L$ and $p:X\to Y$ morphisms in $\mathbf{sSets}$, by the universal property of $\mathbf{Hom}(K,X)\times_{\mathbf{Hom}(K,Y)}\mathbf{Hom}(L,Y)$, commutative diagrams of the form $$ \require{AMScd} \begin{CD} \Lambda^{n}_{k} @>\lambda>>\mathbf{Hom}(L,X)\\ @V\eta VV @V(i^*,p_*)VV \\ \Delta^n @>\sigma>> \mathbf{Hom}(K,X)\times_{\mathbf{Hom}(K,Y)}\mathbf{Hom}(L,Y) \end{CD} $$ are in correspondence with triples of commutatives squares of the form : $$ \require{AMScd} \begin{CD} \Lambda^{n}_{k} @>\lambda>>\mathbf{Hom}(L,X)\\ @V\eta VV @Vi^*VV \\ \Delta^n @>\sigma_1>> \mathbf{Hom}(K,X) \end{CD} $$ $$ \require{AMScd} \begin{CD} \Lambda^{n}_{k} @>\lambda>>\mathbf{Hom}(L,X)\\ @V\eta VV @Vp_*VV \\ \Delta^n @>\sigma_2>>\mathbf{Hom}(L,Y) \end{CD} $$ $$ \require{AMScd} \begin{CD} \Delta^{n} @>\sigma_2>>\mathbf{Hom}(L,X)\\ @V\sigma_1 VV @Vi^*VV \\ \mathbf{Hom}(K,X) @>p_*>> \mathbf{Hom}(K,Y). \end{CD} $$ By the exponential law, triples of commutative squares of this form are in correspondence with triples of commutative squares of the form : $$ \require{AMScd} \begin{CD} \Lambda^{n}_{k}\times K @>\eta \times id_K>>\Delta^n \times K\\ @Vid_{\Lambda^{n}_k}\times i VV @V\tilde{\sigma_1}VV \\ \Lambda^{n}_k\times L @>\tilde{\lambda}>> X \end{CD} $$ $$ \require{AMScd} \begin{CD} \Lambda^{n}_{k}\times L @>\tilde \lambda>>X\\ @V\eta\times id_L VV @VpVV \\ \Delta^n\times L @>\tilde{\sigma_2}>>Y \end{CD} $$ $$ \require{AMScd} \begin{CD} \Delta^{n}\times K @>\tilde{\sigma_1}>>X\\ @Vid_{\Delta^n}\times i VV @VpVV \\ \Delta^n\times L @>\tilde{\sigma_2}>> Y, \end{CD} $$ where $\tilde \lambda$, $\tilde{\sigma_1}$ and $\tilde{\sigma_2}$ are the decurryfied of $\lambda$, $\sigma_1$ and $\sigma_2$ respectively. The commutativity of the first square of the first triple is equavelent to the commutiativity of the first square of the second triple, and so on and so forth. By the universal property of $\Lambda^{n}_{k}\times L\cup_{\Lambda^{n}_k\times K}\Delta^n\times K $, the triples of the second form are in turn in correspondence with commutative squares of the form, $$ \require{AMScd} \begin{CD} \Lambda^{n}_{k}\times L\cup_{\Lambda^{n}_k\times K}\Delta^n\times K @>(\tilde{\sigma_1}, \tilde \lambda) >>X\\ @V(id_{\Delta^n}\times i, \eta \times id_K )VV @VpVV \\ \Delta^n\times L @>\tilde{\sigma_2}>> Y. \end{CD} $$