Inverses of homotopic paths are homotopic too 
PROVE: If $g_1 \simeq g_2$, then $\bar{g_1} \simeq \bar{g_2}$.

I found the solution online here, which I copy it down here for convenience:
(1) Let $\bar{g_1} : I \to X$ be defined by $\bar{g_1}(s) = g_1(1-s)$, and similarly $\bar{g_2}(s) = g_2(1-s)$ 
(2) Let $H(s, t)$ be a homotopy from $g_1$ to $g_2$ so that $H(s, 0) = g_0(s)$ and $H(s, 1) = g_1(s)$ 
(3) Define $\bar{H}(s, t) = H(1 - s, t)$ 
(4) The map $(s, t) \to (1 - s, t)$ is a homeomorphism from $I \times I$ to itself, so we see that $\bar{H}(s, t)$ is a homotopy from $\bar{g_1}(s)$ to $\bar{g_2}(s). \quad \blacksquare$
Do you have alternative proof that is more intuitive than the above? I especially found (4) un-intuitive, can you explain it in differently? Thanks for your time and effort.

PS: This question looks very classic, but I am surprised that it was never asked before. The only closest sibling similar to mine is here, but they are not the same. The same question was raised here but the proof was not given. Let me know if you found any.
 A: The key in (4) is that $(s, t) \mapsto (1 - s, t)$ is continuous which follows from the fact that $s \mapsto 1 - s$ is continuous.
Notice what the domain and codomain of $H$ are, that is, $H : I \times I \to X$. We defined $\overline H : I \times I \to X$ by $\overline H(s, t) = H(1 - s, t)$. We can define $\varphi : I \times I \to I \times I$ by $\varphi(s, t) = (1 - s, t)$ which we've just established is a continuous map. Notice that $$\overline H(s, t) = H(1 - s, t) = H\big( \varphi(s, t) \big).$$
or in the form of a commutative diagram:
$\hspace{7cm}$
Then $\overline H$ is a composition of $H$ with a continuous function. Since $H$ is a homotopy by hypothesis, it is continuous, so the composition is continuous.
Now just notice that $$\overline H(s, 0) = H(1 - s, 0) = g_1(1 - s) = \overline g_1(s)$$ and similarly $$\overline H(s, 1) = H(1 - s, 1) = g_2(1 - s) = \overline g_2(s)$$ which tells you $\overline H$ is a homotopy between $\overline g_1$ and $\overline g_2$.
The person who wrote the original proof added more information than needed: they said $(s, t) \mapsto (1 - s, t)$ was a homeomorphism (which is true), when all we needed was continuity.
