I've come across the following claim here, in lemma (1.15): if $X$ and $Y$ are simplicial sets and $Y$ is a Kan complex, then for any $k\leq n$ and $\varphi:X\times\Lambda^n_k\rightarrow Y$ there is a simplicial morphism $\tilde\varphi:X\times\Delta^n\rightarrow Y$ making the triangle below commute:
The proof given in that book is not fleshed out at all and I have been struggling to fill in the details. Using the simplicial extension theorem one can show the result fairly easily, as if $\lvert\cdot\rvert:\textbf{Set}^{\Delta^{op}}\rightarrow\textbf{Top}$ is the realization functor then $\lvert X\times\Lambda^n_k\rvert\cong\lvert X\rvert\times\lvert\Lambda^n_k\rvert$, and using any retraction $r:\lvert\Delta^n\rvert\rightarrow\lvert\Lambda^n_k\rvert$ we have that this triangle commutes:
I suspect there is also probably a proof using the internal $\textbf{Hom}$ in $\textbf{Set}^{\Delta^{op}}$ although I have not thought this through. In any case, these seem like overkill to me, as the proof given in the linked pdf is strictly combinatorial, and I would very much like to figure out the purely combinatorial proof to get better intuition for the combinatorics of Kan complexes. Unfortunately I have been struggling with this. Here is my approach thus far and the fundamental issue with it:
Throughout denote face maps by $d_i$ and degeneracy maps by $s_i$. We wish to define $\tilde\varphi$ inductively on the skeletons of $X\times\Delta^n$. In the base case, for $i<n-1$, we have $\Delta^n_i=(\Lambda^n_k)_i$, so let $\tilde\varphi_i=\varphi_i:X_i\times\Delta^n_i\rightarrow Y_i$.
Now we need to define $\tilde\varphi_{n-1}:X_{n-1}\times\Delta^n_{n-1}\rightarrow Y_{n-1}$. Let $\sigma\in X_{n-1}$. If $E^n\in\Delta^n_n $ is the one nondegenerate $n$-simplex of $\Delta^n$, then for any $n-1$-simplex $\tau\neq d_k E^n\in \Delta^n_{n-1}$ we have $\tau\in(\Lambda^n_k)_{n-1}$, so we must have $\tilde\varphi_{n-1}(\sigma, \tau)=\varphi_{n-1}(\sigma, \tau)$. Hence we need only define the $\tilde\varphi_{n-1}(\sigma, d_k E^n)$. I could not think of a canonical way of doing this, but I realized that by naturality of $\tilde\varphi$ we must have $\tilde\varphi_{n-1}(\sigma, d_k E^n)=d_k \tilde\varphi_{n-1}(s_k \sigma, E^n)$, so it would suffice to define $\tilde\varphi_n:X_n\times\Delta^n_n\rightarrow Y_n$ and then apply a face map.
Hence let $\sigma\in X_n$. For $\tau\in\Delta^n_n$ such that $\tau\neq s_l d_k E^n$ for any $l\in\{0, ..., n-1\}$, we again have $\tau\in(\Lambda^n_k)_n$, and so we must have $\tilde\varphi_n(\sigma, \tau)=\varphi_n(\sigma, \tau)$. Thus we need only define $\tilde\varphi_n(\sigma, s_l d_k E^n)$ for each $l$. First consider the case $l\in \{k, k+1\}$. Then $s_l d_k E^n=E^n$, so we wish to define $\tilde\varphi_n(\sigma, E^n)$. For this we will use that $Y$ is a Kan complex; indeed, for all $j\in\{0, ..., n\}$ with $j\neq k$, define $y_{\sigma, j}=\tilde\varphi_{n-1}(d_j\sigma, d_j E^n)=\varphi_{n-1}(d_j\sigma, d_j E^n)\in Y_{n-1}$. Then for all $i<j\in\{0, ..., n\}$ with $i, j\neq k$, we have:
$\begin{align} d_i y_{\sigma, j} &= d_i \varphi_{n-1}(d_j\sigma, d_j E^n)=\varphi_{n-2}(d_i d_j\sigma, d_i d_j E^n)\\\\&=\varphi_{n-2}(d_{j-1} d_i\sigma, d_{j-1} d_i E^n)=d_{j-1}\varphi_{n-1}(d_i\sigma, d_i E^n)=d_{j-1}y_{\sigma, i}\end{align}$
Hence the $y_{\sigma, j}$ satisfy the relevant face matching conditions and so, since $Y$ is fibrant, there is an $n$-simplex $y_\sigma\in Y_n$ such that $d_j y_\sigma=y_{\sigma, j}$ for all $j\neq k$. Thus define $\tilde\varphi_n(\sigma, E^n)=y_\sigma$. All of this seems fairly canonical to me.
However, this is where my problem arises; it is not clear to me that this choice is well-defined. In particular, I don't see how this construction guarantees compatibility along different $n$-simplices of $X$; ie that if $\sigma^1, \sigma^2\in X_n$ have $d_k \sigma^1=d_k \sigma^2$, we will have $d_k y_{\sigma^1}=d_k y_{\sigma^2}$. Certainly if $d_j \sigma^1=d_j \sigma^2$ for $j\neq k$, we will have $d_j y_{\sigma^1}=y_{\sigma^1, j}=\tilde\varphi_{n-1}(d_j\sigma^1, d_j E^n)=(d_j\sigma^2, d_j E^n)=y_{\sigma^2, j}=d_j y_{\sigma^2}$, but I do not see how we can possibly ensure that the $k$-th faces will coincide. In fact, if the $y_\sigma$ are chosen as arbitrarily as above, I imagine that in general the $k$-th faces will not coincidence, eg by considering a case where $Y$ is the singular simplex of some open ball in $\mathbb{R}^2$, $n=2$, and $X$ is a simplicial complex with precisely two nondegenerate $2$-simplices, whose only common face is their $k$-th one. (If I have time I will try to illustrate this with drawings and upload the photos later.)
If we could resolve this problem it seems fairly clear to me how we might proceed to define $\tilde\varphi_n$ on the rest of $X_n\times\Delta^n_n$, and then inductively continue up to define $\tilde\varphi$ on the the higher skeleta $X_m\times\Delta^n_m$ for $m>n$, at every stage using that $Y$ is fibrant and a similar kind of construction and compatibility argument. Unfortunately I just don't see what the compatibility argument could be; does anyone have any insight on this? Alternatively, does anyone have an alternative approach to a combinatorial proof of this fact? Everything in my approach thus far seems to have been forced, but perhaps I am missing something key. Sorry for the extremely long post; even just hints, tips, or ideas would be greatly appreciated. Thank you so much in advance.