You can convert the diagonal $\Delta: X \rightarrow X \times X = X^2$ to a pointwise homotopy equivalent fibration $f:E \rightarrow B$ by letting $B = X \times X$ and $E = X^I$. If $\gamma : I \rightarrow X$ is a path in $X$ the map $f$ is given by $f(\gamma) = (\gamma(0),\gamma(1))$.
Alright but since $I$ only has two endpoints I don't see how to generalize this result to powers of $X$ higher than $2$.
What if we have the diagonal $\Delta: X \rightarrow X \times ... \times X = X^i$ for some finite $i$? How do we convert $\Delta$ to a fibration with $B = X^i$? I was thinking we let $J$ be the space obtained by gluing together $i$ copies of the interval $I$ at $1$ and then let $E = X^{J}$. There are $i$ maps $* \rightarrow J$ given by inclusion at all the zeroes of the intervals in $I$ which induce $i$ evaluation maps $X^J \rightarrow X$, so a map $f:X^J \rightarrow X^i$. This map is clearly pointwise homotopy equivalent to $\Delta$ but I can't prove that it is a fibration.
Thanks for all the help in advance! :)
Edit:
I guess this can be proved by showing that if $A \rightarrow B$ is the inclusion of a subspace then the restriction map $X^B \rightarrow X^A$ is a fibration for suitable spaces, maybe when $A \rightarrow B$ is a cofibration?