Homotopy equivalences in pushout square with cofibration.

Question

If following square is a pushout square, $g$ is a cofibration and $f$ is a homotopy equivalence then $i$ is also a homotopy equivalence. $$ \begin{matrix} A & \xrightarrow{f} & B \\ \left\downarrow{g}\vphantom{\int}\right. & & \left\downarrow{h}\vphantom{\int}\right.\\ C& \xrightarrow{i} & D \end{matrix} $$

My idea was to take firstly the homotopy $H$ which makes $ f \circ \tilde f \simeq \mathrm{id}_B$ and then write down arrows $g \circ H(\square,t):B \to C$ and $\mathrm{id}_C:C\to C$ from universal property of push-out we'd obtain arrow $i_t:D\to C$ then we could "glue" homotopy equivalence between $i_1 \circ i \simeq \mathrm{id}_C$, by formula $\tilde H(c,t) = i_t(c)$ but we don't know whether it is continuous function in $t$. Is there any way to show that (this "induced" homotopy $\tilde H$ seems to be unique so intuition tells me that it is possible). And how to show existence of $\tilde f \circ f \simeq \mathrm{id}_A$ homotopy? I'd be grateful for hints and answers.

Answer 1

An answer to this is that the result required is a special case of the gluing theorem for homotopy equivalences discussed mathoverflow and proved in Topology and Groupoids. The relevant Chapter 7 on "Cofibrations" is available here. This result was found, as explained there, by generalising the proof that a homotopy equivalence of spaces induces an isomorphism of homotopy groups, and so the proof given there has the advantage of giving control of the homotopies involved.