This is probably a simple question but I can't seem to find any references for it. Call a Kan complex $X$ $k$-truncated if $\pi\_n(X, x) = 0$ for all $x \in X\_0$ and $n > k$.


Claim: $X$ is $k$-truncated iff $\text{Fun}(S,X)$ is $k$-truncated for every simplicial set $S$, where $\text{Fun}(S,X) = X^S$ is the mapping space.


One direction is obvious, take $S = \Delta^0$, then $\text{Fun}(\Delta^0, X) \cong X$. So if $\text{Fun}(S,X)$ is $k$-truncated for all $S$, then $X$ is $k$-truncated.


How do I prove the other direction?



A space $X$ is $k$-truncated iff $\text{Fun}(S,X)$ is $k$-truncated
Let $S^n := \Delta^n/\partial \Delta^n$; and let me assume for simplicity that $X$ is connected.


We have a (homotopy) fiber sequence $\Omega^n X \to X^{S^n} \to X$.


In particular, for $n>k$, $\Omega^n X$ is contractible (it is a Kan complex and its homotopy groups are trivial by assumption), so that for $n>k$, $X^{S^n}\to X$ is a homotopy equivalence (it is an equivalence on homotopy groups by the long exact sequence, and they are both Kan complexes)


It follows that $(X^{S^n})^S\to X^S$ is also a homotopy equivalence, so that $(X^S)^{S^n}\to X^S$ is also a homotopy equivalence.


Taking any $f\in X^S$ and the associated fiber sequence, we find that $\Omega^n(X^S,f)$ is also contractible, and thus $X^S$ is $k$-truncated.


