Denote $[n] := \{0,\dots,n \}$. It is well ordered set. There is a fact: $$ \hom([n-1],[1]) \simeq [n],$$ where elements of $\hom$ are nondecreasing maps.
This result has an easy explanation: nondecreasing map between $[n-1]$ and $[1]$ is the same thing as to place one bar in $[n-1] =\{0,\dots,n-1\}$. There is an obvious order with this bar: the more to the left the bar the more corresponding element is.
Let's consider an injection $[k-1] \hookrightarrow [n-1] $. So we have the map $$\hom([n-1],[1]) \to \hom([k-1],[1]) $$ which is by the fact above is
$$[n] \to [k].$$
You can easily check that this map is surjective. So we have constructed $$[n] \twoheadrightarrow [k]. $$
It is also easy to check that this construction is bijection between injections $[k-1] \hookrightarrow [n-1] $ and surjections $[n] \twoheadrightarrow [k]$, where $n,k$ are fixed.
My question is about inverse map. I showed how we can construct surjective map from injective: you need to apply $\hom(-,[1])$. How to explicitly construct injective map $[k-1] \hookrightarrow [n-1] $ from surjection $[n] \twoheadrightarrow [k]$?
If we have surjection $[n] \twoheadrightarrow [k]$, by the first fact above we have $$\hom([n-1],[1]) \twoheadrightarrow \hom([k-1],[1]).$$ I don't know what to do the next. Maybe I should apply Yoneda lemma, but I don't know how.
I will appreciate any hints!
I use $\Delta$ as standard notation for the category of finite ordinals. I assume some basic familiarity with the notation of simplicial sets.
I'd like to pedantically point out that this is not immediately true. So far you've only explained a bijection of sets: $$\hom([n-1],[1])\cong[n]$$But have given no indication as to whether or not this bijection is natural. What's more, to even make it make sense as "natural", we'd need to provide some kind of contravariant functor $\Delta\to\mathsf{Set}$ which maps $[j]$ to $[j+1]$ and has some kind of action on arrows. To go further and claim that we get an order preserving surjection $[n]\to[k]$ from an order preserving injection $[k-1]\to[n-1]$, you'd need to make sure this functor also preserves the order i.e. lifts to a contravariant functor $\Delta\to\Delta$. To go even further and claim arrows $[n]\to[k]$ in $\Delta$ are the same as arrows $[k-1]\to[n-1]$ you'd need the (contravariant) lift $\Delta\to\Delta$ to be fully faithful (or at least, fully faithful when the codomain is a smaller ordinal than the domain).
Define a simplicial set $X$ through $X_j=[j+1]$ and by the face maps $d_i:X_j\to X_{j-1},x\mapsto\sigma_i(x)$ and the degeneracies $s_i:X_{j-1}\to X_j,x\mapsto\delta_{i+1}(x)$. $X$ obviously lifts to a contravariant functor $\Delta\to\Delta$ with the desired properties. So now we can more precisely ask whether or not the following simplicial sets are isomorphic: $$\hom(-,[1])\overset{?}{\cong}X$$And indeed they are, by the same isomorphism you describe, but very importantly we can now discuss its naturality (exercise).
What if we want to keep things explicit? Fix $k,n\in\Bbb N_0$. Say $f:[k]\to[n]$ is a monotone injection. We want to create a monotone surjection $[n+1]\to[k+1]$. That's not too hard: $$g:[n+1]\ni j\mapsto\begin{cases}\min\{i\in[k]:f(i)\ge j\}&0\le j\le f(k)\\k+1&f(k)<j\le n+1\end{cases}$$The injectivity of $f$ makes $g$ surjective, and $g$ inherits monotonicity from $f$ too. In the other direction, given a monotone surjection $g:[n+1]\to[k+1]$ we can define: $$f:[k]\ni j\mapsto\max\{i\in[n]:g(i)=j\}\in[n]$$Which is obviously again monotone and an injection. The processes $f\leftrightarrow g$ are mutually inverse, so, we are happy.
N.B. if we drop the requirement of being order preserving it's no longer true that injections $[k]\to[n]$ are in correspondence with surjections $[n+1]\to[k+1]$.