Suppose we have category $\mathcal{C}$ which has the structure of a category of fibrant objects, and suppose we have a functor $F:I\to \mathcal{C}$ with a limit $\lim F$ in $\mathcal{C}$.
If we have two maps $t_1,t_2:T\to \lim F$ such that for every $\pi_i:\lim F \to F_i$ satisfies $\pi_i\circ t_1\simeq\pi_i\circ t_2$, can we show that the two maps $t_1$ and $t_2$ are homotopic?
My first idea was to consider the localisation functor $\gamma:\mathcal{C}\to Ho(\mathcal{C})$, in the naive hope of it preserving finite limits, which it does not necessarily do unfortunately. However, intuitively it should be possible to use the homotopies $\eta_i:\pi_i\circ t_1\to \pi_i\circ t_2$, using the notation of Definition 4.16 on the nCatLab page, to assemble this into a homotopy $\eta:t_1\to t_2$.
Does anyone have any idea if this is true, or not? And, possibly give a reference to existing literature, or give a small argument on why it (does not) holds? Or, maybe a characterization of cases in which this does hold? For example; if all maps in the diagrams in $\mathcal{C}$ are weak equivalences.
Thanks!
The category of topological spaces with the Quillen Model Structure is a category of fibrant objects (it really doesn't matter which structure you chose, as long as all objects are fibrant). Consider the (limit) pullback diagram [
Both $\mathbb{R}^2$ and the point $\ast$ are contractible so that any two maps $S^1 \rightarrow \mathbb{R}^2, \ast$ are homotopic. Since $\pi_1(S^1) = \mathbb{Z}$ we know that there are non homotopic maps $t_1$ and $t_2$ such that the composition with the legs of the cone are homotopic.
I don't know much about categories of fibrant objects, but in a model category weak equivalences are ismomorphisms in the homotopy category, so if every map and object in the diagram were fibrant-cofibrant and and least one leg $\pi_i$ were a weak equivalence we would have the following:
$\pi_i \circ t_1 \cong \pi_1 \circ t_2$ then $[\pi_i \circ t_1] = [\pi_1 \circ t_2]$, here [-] denote homotopy classes of morphisms. Rewriting, $[\pi_i] \circ [t_1] = [\pi_i] \circ [t_2] $ and since $[\pi_i]$ is an isomorphism we would have that $[t_1]=[t_2]$