Why does here shifting by $+ \tau$ on one side implies shifting by $- \tau$ on the other? I don't understand a part of the solution of an exercise and would be grateful if you could help.

Note: $\psi(t)$ and $x(t)$ are time continuous signals (german: "zeitkontinuierliche Signale"). The signal $\psi(t)$ is the same as in the Wavelet Transform https://en.wikipedia.org/wiki/Continuous_wavelet_transform
The follwing equality needs to hold: 
$\int_{-\infty}^{\infty} x(\tau)\frac{1}{a}\psi (\frac{\tau -t}{a}) d\tau \overset{!}{=} \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau$
So $h(t-\tau) \overset{!}{=} \frac{1}{a}\psi (\frac{\tau -t}{a})$, i.e. $h(t) \overset{!}{=} \frac{1}{a}\psi (\frac{-t}{a})$

My question is about the last equality, $h(t) \overset{!}{=} \frac{1}{a}\psi (\frac{-t}{a})$. Why does shifting $h(t-\tau)$ by $+ \tau$ implies shifting $\psi (\frac{\tau -t}{a})$ by $-\tau$ and not $+\tau \ $?
Thanks for you help !

Because your emphasis is on "shifting", I'm not entirely sure what you are asking. But if $u=t-\tau  $, then  $\tau-t=-u $. It's just that substitution. 
