Derivation of Lorentz Factor yielding $\gamma ^{-1}$ as opposed to $\gamma$ I just wish to understand why the following reasoning fails.
Suppose observer $B$ (moving reference frame) is moving at a relative velocity $v$ from observer $A$ (stationary reference frame) along the $x$-axis. Observer $B$ shines a ray of light along the $y$-axis, and the ray travels $1m$ before hitting a wall stationary relative to $B$.
From $B$'s point of view, the ray took $t_B=c^{-1}\ \text{seconds}$ to reach the wall, while from $A$'s point of view the ray took $$t_A=\frac{1}{\sqrt{c^2-v^2}}\ \text{seconds}$$
to reach the wall. The above formula can be derived by drawing a right-angle triangle with hypotenuse $ct_A$ and the other sides given by $1m$ and $vt_A$. Then one only needs to solve for $t_A$ in the equation
$$(ct_A)^2=(vt_A)^2+1.$$
This yields, however, that
$$t_B=\gamma ^{-1}t_A$$

Which step in the derivation is wrong?
 A: Your derivation is correct and one further step gives $$t_A =\gamma t_B,$$ completing the derivation. Though the $t_B$ is not quite a proper time, it is half of a different proper time, which would hold if the wall had a mirror which reflected the light back to the emitter of $B$.
Times are always dilated relative to the proper time. A time is defined between two events in spacetime that are timelike-separated, and it turns out that when things are objectively time-separated in relativity they are not objectively space-separated, so there is always a reference frame that thinks both happened at the same point in space, and they measure the proper time—everyone else measures something longer.
You have concocted a scenario where the two events are null-separated, this technically means that the events are objectively both space and time separated, but those separations can be driven arbitrarily close to 0 by choice of reference frame. But, you have mercifully forbidden us from accelerating in the $y$-direction, so thankfully these details can be avoided, and to the easiest way to accomplish this is to make the problem one-dimensional by reflecting the light back at the $x$-axis using the Parity transform, adding these two null-vectors creates a timelike four-vector.
