Can anyone squash KaTeX bugs?

Both Checked on tiddlywiki.com with katex installed.

When using the KaTeX \begin{align} environment with fixed story fluid sidebar the text does not wrap to the river but sticks out past the boundary.

The second appears to be a my wiki/ css problem.

Using the \begin{align} environment should number the equations down the right hand side.

on tiddlywiki.com this happens correctly, ie (1) (2) (3) against each equation.
On my wiki i get (1) (1) (1)…

The issue appears to be with @telmiger 's stylesheet for textstretch.
Can be found here
Deleting the ‘Foot notes and numbering’ section of the css seems to fix the katex problem.

/* * * * * * * * * * * *
** Footnotes with Numbers
* * * * * * * * * * * * */

body {
   counter-reset: notenr;  /* set counter to 0 */
}
div .tc-tiddler-frame {
   counter-reset: tidnotenr;
}
.strex-container.storynumbers {
   counter-increment: notenr; /* counter +1 */
}
.strex-container.numbers {
   counter-increment: tidnotenr;
}
button.strex-open.storynumbers::before, 
button.strex-start.storynumbers::before {
   content: counter(notenr); /* Display the counter */
   font-size: xx-small;
   vertical-align: top;
}
button.strex-end.storynumbers::after {
   content: counter(notenr);
   font-size: xx-small;
   vertical-align: top;
}
button.strex-open.numbers::before, 
button.strex-start.numbers::before {
   content: counter(tidnotenr);
}
button.strex-end.numbers::after {
   content: counter(tidnotenr);

(Engineer's Text Book — A Hyperlinked Guide For Engineers)
Any thoughts anyone?
@pmario @jeremyruston

My wiki: Engineer's Text Book — A Hyperlinked Guide For Engineers

Test code:

$$\begin{align}
	\frac{B}{T_1}-\frac{B}{T_2}&=(\ln (\mu_1)-\ln (D))-(\ln (\mu_2)-\ln (D)) \\
	\text{a }-,-= + \text{ so:} \\
	\frac{B}{T_1}-\frac{B}{T_2}&=\ln (\mu_1)-\ln (D)-\ln (\mu_2)+\ln (D) \\
	\text{This means the } \ln(D) \text{ drops out} \\
	\frac{B}{T_1}-\frac{B}{T_2}&=\ln (\mu_1)-\cancel{\ln(D)}-\ln (\mu_2)+ \cancel{\ln(D)} \\
	\frac{B}{T_1}-\frac{B}{T_2}&=\ln \mu_1- \ln \mu_2 \\
	\text{at this point it is worth noting that: }
	\frac{B}{T}=BT^{-1} \\
	BT^{-1}_1-BT^{-1}_2&=\ln \mu_1- \ln \mu_2 \\
	\text{Putting }T \text{ into brackets:} \\
	B(T_1^{-1}-T_2^{-1})&=\ln \mu_1- \ln \mu_2 \\
	B&=\frac{\ln \mu_1- \ln \mu_2}{T_1^{-1}-T_2^{-1}}
\end{align}$$