I was wondering if there is an analogous of the Problem 8.49 from Strom, Modern classical homotopy theory in a Model Category.



> 
> There is a bijection between $[X,Y]$ and $<X,Y>$ when Y is a simply connected space.
> 
> 
> 


Where


1. $[X,Y]$ are pointed homotopy classes of maps $X\to Y$;
2. $\langle X,Y\rangle$ are free homotopy classes of maps $X\to Y$.



Analogous of Strom 8.49 for Model Categories
Say $\mathcal{M}$ is a model category and $A$ is a cofibrant object of $\cal{M}$. Under these conditions, there's an undercategory $\mathcal{M}\_{A/}$ whose objects are pairs of an object $X$ of $\mathcal{M}$ and a map $A \to X$, and whose morphisms are commutative triangles. The undercategory has a model structure, where maps are cofibrations, fibrations, or weak equivalences iff the same is true after forgetting about the map from $A$. If $\mathcal{M}$ is spaces and $A$ is a point, this undercategory is the category of pointed spaces. So we could ask: given objects $X$ and $Y$ under $A$, when is the "forgetful map" $[X,Y]\_{A/} \to [X,Y]$ guaranteed to be an isomorphism?


Since we're computing maps in the homotopy category, we might as well assume $Y$ is fibrant in $\mathcal{M}\_{A/}$ (which is the same as $Y$ being fibrant in $\mathcal{M}$) and $X$ is cofibrant in $\mathcal{M}\_{A/}$ (which is the same as $A \to X$ being a cofibration) -- which implies that $X$ is cofibrant in $\mathcal{M}$ too because $A$ is cofibrant.


For the following we'll assume that $M$ is a simplicial (or topological) model category so that we can talk about mapping spaces; then cofibrance-fibrance implies:
$$
\begin{align\*}
[X,Y] &= \pi\_0 Map\_{\mathcal{M}}(X,Y)\\
[X,Y] &= \pi\_0 Map\_{\mathcal{M}\_{A/}}(X,Y)
\end{align\*}
$$
The axioms for a simplicial model category (particularly SM7) also tell us that the map
$$
Map\_{\mathcal{M}}(X,Y) \to Map\_{\mathcal{M}}(A,Y)
$$
is a fibration, whose fiber over the reference map $A \to Y$ is $Map\_{\mathcal{M}\_{A/}}(X,Y)$. Therefore, we can apply the long exact sequence on homotopy groups and get
$$
\dots \to \pi\_1 Map\_{\mathcal{M}}(A,Y) \to [X,Y]\_{A/} \to [X,Y] \to [A,Y].
$$


Therefore, in this circumstance, you always get a bijection between "free" homotopy classes and homotopy classes under $A$ when the mapping space $Map\_{\mathcal{M}}(A,Y)$ is 1-connected.


I'd hope that there's an argument that doesn't use mapping spaces, but it would be harder.


