Let $A_0\xrightarrow{f_0} A_1\xrightarrow{f_1} A_2\xrightarrow{f_2}\cdots$ be a system of cofibrations in which all the $A_i$ are contractible. How do I prove that $colim_iA_i$ is also contractible?
I so far have this.
We know the diagram $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} A_n&\ra{f_n}&A_{n+1}\\ \da{i_0}&&\da{}\\ A_{n}\times I&\ra{}&Cylf_n \end{array} $$ exists for all $n\in\mathbb{N}$ where $Cylf_n$ is the mapping cylinder of $f_n$. We also know the diagram $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} A_n&\ra{f_n}&A_{n+1}\\ \da{i_0}&&\da{i_0}\\ A_{n}\times I&\ra{f_n\times id_I}&A_{n+1}\times I \end{array} $$ exists. Since $Cylf_n$ is a pushout, there exists a unique map $\alpha_n:Cylf_n\to A_{n+1}\times I$ that makes the appropriate diagram commute. Since $f_n$ is a cofibration, there exists a continuous function $\beta_n:A_{n+1}\times I\to Cylf_n$ that is a left inverse for $\alpha_n$ which means that $\beta_n\alpha_n=id_{Cylf_n}$.
Now since each $A_n$ is contractible, there exists continuous functions, $g_n:A_n\to *$ and $h_n:*\to A_n$ such that $g_n\circ h_n\sim id_*$ and $h_n\circ g_n\sim id_{A_n}$ for all $n\in\mathbb{N}$. This means there exists homotopies $H_n:*\times I\to *$ and $\hat H_n:A_n\times I\to A_n$ such that $H_n(*,t)=*, \hat H_n(x,0)=x,$ and $\hat H_n(x,1)=g_n(h_n(x))$ for all $n\in\mathbb{N}$.
I don't know how to proceed from here.