The inclusion of a space into the colimit is a cofibration. 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 each inclusion of a space into the colimit is a cofibration?
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 $n\in\mathbb{N}$.
Now consider the inclusion $i_n:A_n\to colim_iA_i$.
We know the diagram
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
A_n&\ra{i_n}&colim_iA_i\\
\da{i_0}&&\da{}\\
A_{n}\times I&\ra{}&Cyli_n
\end{array}
$$
exists for all $n\in\mathbb{N}$ where $Cyli_n$ is the mapping cylinder of $i_n$.
We also know the diagram 
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
A_n&\ra{i_n}&colim_iA_i\\
\da{i_0}&&\da{i_0}\\
A_{n}\times I&\ra{i_n\times id_I}&colim_iA_i\times I
\end{array}
$$
exists.
Since $Cyli_n$ is a pushout, there exists a unique map $\gamma_n:Cyli_n\to colim_iA_i\times I$ that makes the appropriate diagram commute.
I need a continuius left inverse to $\gamma_n$
 A: I'm not sure what you're doing in your work in the question. 
Let $A_\infty$ be the colimit. We want to prove that $A_i\to A_\infty$ is a cofibration. We'll directly show that it has the homotopy extension property.
Let $f: A_\infty\to X$ be a map, $H:A_i\times I \to X$ a homotopy with $H|_{A_i\times \{0\}} = f|_{A_i}$.
Let $H_i=H$, and then inductively choose $H_n :A_n\times I\to X$ to be an extension of
the homotopy $H_{n-1}$ that agrees with $f|_{A_n}$ at time $t=0$.
Since the interval is locally compact Hasudorff, $-\times I$ has a right adjoint 
$(-)^I$, and we can regard $H_n$ as being 
equivalent to a map $A_n\to X^I$, where $X^I$ is the space of continuous functions from $I$ to $X$ with the compact-open topology. Let $H_n':A_n\to X^I$ be the corresponding function, defined by
$$H_n'(a) = t\mapsto H_n(a,t).$$
By construction the $H_n'$s form a cocone out of the sequence of $A_n$s. (Define $H_n$ for $n<i$ by just restricting $H$). Thus the $H_n'$s induce a canonical map 
$H_\infty' : A\to X^I$, which we can adjoint back over to get a map 
$H_\infty : A\times I \to X$.
We have $$H_\infty(a_n,0) = H_\infty'(a_n)(0) =  H_n'(a_n)(0)=H_n(a_n,0)=f(a_n),$$
where $a_n$ is the image of a point of $A_n$ in $A_\infty$. Thus $H|_{A_\infty\times\{0\}}=f$, as desired.
Additionally, if $a\in A_i$, then 
$$H_\infty(a,t) = H_\infty'(a)(t)=H_i'(a)(t)=H_i(a,t)=H(a,t).$$
Thus $H_\infty$ also extends $H$, as required.
Thus $A_i\to A_\infty$ is a cofibration.
