I'm trying to prove that given a continuous map $f : X \to Y$, the inclusion
$$ i : x \in X \mapsto [(x,1)] \in M_f $$
is a cofibration.
Given a homotopy $H : X \times I \to W$ and $g : M_f \to W$ so that
$$ H(x,0) = gi(x) = g([(x,1)]) $$
we want a homotopy $J : M_f \times I \to W$ such that for all $x \in X$ we have
$$ J([(x,1)],t) = H(x,t) \text{ and } J([(x,0)],t) = g(f(x)). $$
Since $g$ is a function with domain $M_f$, which is a pushout, we have continuous functions $g_X : X \times I \to W$ and $g_Y : Y \to W$ such that $$ g_Y(y) = g([y]), \quad g_X(x,t) = g([(x,t)]), $$
and $g_X(x,0) = g_Y(f(x))$.
Now, since $I$ is locally compact Hausdorff, by the exponential law the endofunctor $$- \times I : \mathsf{Top} \to \mathsf{Top}$$ is left adjoint to $\mathsf{Top}(I,-) :\mathsf{Top} \to \mathsf{Top}$. Hence the former preserves colimits and so $M_f \times I$ is the pushout of $X \times I \xrightarrow{i_0 \times 1} (X \times I) \times I$ and $X \times I \xrightarrow{f \times 1} Y \times I$. Thus, the map $J$ corresponds to a certain selection of maps $\tilde{H} : (X \times I) \times I \to W$ and $\tilde{g} : Y \times I \to W$ so that $$\tilde{g}(f(x),t) = \tilde{H}(x,0,t).$$
My intuition here is to leave the 'initial data' $g$ fixed on $Y$, and at each instant $s$ to have a function from $X \times I$ that behaves like $H_0$ in the bottom face and coinides with $H_s$ at the top face. This would make $J$ an extension of the original homotopy. Hence I wanted to define
$$ \tilde{H}(x,s,t) = H(x,ts) \text{ and } \tilde{g}(y,t) =g_Y(y). $$
However, for these to induce a map from $M_f$ we should have that $g([(x,0)]) = g([(x,1)])$ for all $x \in X$. But moreover, for the induced map to preserve the initial data, one should have $g = Ji_0$ and $H = J(i \times 1_I)$ which amounts to proving
$$ g([(x,t)]) = g([(x,0)]) \quad (\forall t \in I). $$
It is not evident to me that this is the case. What's missing here? Are my choices for $\tilde{H}$ and $\tilde{g}$ incorrect? Any help is much appreciated.
