The problem is the following: Suppose $X_0\rightarrow X_1\rightarrow\cdots$ is a sequence of cofibrations. We will denote $f_n:X_n\rightarrow X_{n+1}$. Assume that we have also maps $v_n:X_n\rightarrow T$ such that $v_n\simeq v_{n+1}\circ f_n$ for each $n\in\mathbb{N}$. We want to show that there exists maps $w_n:X_n\rightarrow T$ such that $w_n=w_{n+1}\circ f_n$. Thus we want to show that we can make strict commutativity out of comutation up to homotopy.
I have done this up to this moment: From the assumption that $v_n\simeq v_{n+1}\circ f_n$ we obtain $H_n:X_n\times I\rightarrow T$ with $H_n(-,0)=v_{n+1}\circ f_n$ and $H_n(-,1)=v_n$. Hence we get a map $\tilde{H}_n:X_n\rightarrow T^I$ by $\tilde{H}_n(x)(t)=H_n(x,t)$. It is obvious that $\varepsilon_0\circ\tilde{H}_n=v_n\circ f_n$ where $\varepsilon_0:T^I\rightarrow T$ is the evaluation in $0$. Since $f_n$ is a cofibration we get a map $h_n:X_{n+1}\rightarrow T^I$ such that $h_n\circ f_n=\tilde{H}_n$ and $\varepsilon_0\circ h_n=v_n$ (Drawing diagrams will be helpful).
But I can't up with a step to construct the maps $w_n$ with right property. Is the idea is described above a good start? Can someone help me to come further?