This is a very general fact about model categories and homotopy pullbacks, as evidenced by Zhen Lin's comment. It's also proven as a special case of Proposition 4.65 in Hatcher's book. Let me nevertheless spell out the argument precisely for topological spaces.
Define $E_p = \{ (y, \gamma) \in E \times B^{[0,1]} \mid p(y) = \gamma(0) \}$. There's a map (in fact a fibration) $q : E_p \to B$, $(y,\gamma) \mapsto \gamma(1)$, and the homotopy fiber is the fiber $F_p = q^{-1}(b_0)$. The inclusion $$i : F = p^{-1}(b_0) \to F_p$$ is given by $i(y) = (y, \mathrm{cst}_{b_0})$.
Define a homotopy $g_t : E_p \to B$, $(y,\gamma) \mapsto \gamma(t)$. Then $g_0(y, \gamma) = \gamma(0) = p(y)$, so $g_0$ lifts through $p$ by $\bar{g}_0 : E_p \to E$, $\bar{g}_0(y,\gamma) = y$ (i.e. $p \circ \bar{g}_0 = g_0$). Because $E \to B$ is a fibration, by the homotopy lifting property, there is a full lift $\bar{g}_t : E_p \to E$ of $g_t$ through $p$. In other words, $\bar{g}_t$ satisfies the following equation:
$$p(\bar{g}_t(y,\gamma)) = \gamma(t).$$
Now restrict everything to the fibers: let $h_t : F_p \to F_p$ be given by $h_t(y,\gamma) = \bigl(\bar{g}_t(y,\gamma), \gamma_{\mid [t,1]} \bigr)$ (because of the previous equation, this is in $F_p$). Then $h_0$ is the identity, whereas $h_1(y,\gamma) = (\bar{g}_1(y,\gamma), \mathrm{cst}_{b_0})$ is in the image of $i : F \to F_p$. And now that $h_t$ is a homotopy between $ih_1$ and the identity, while the restriction of $h_t$ is a homotopy between $h_1i$ and the identity. Thus $F$ and $F_p$ are homotopy equivalent.
Final remark: it's much simpler to prove that $F$ and $F_p$ are weakly homotopy equivalent, because the map $i$ induces easily an isomorphism on all homotopy groups. The square
$$\require{AMScd}
\begin{CD}
E @>>> E_p \\
@VVV @VVV \\
B @>>> B
\end{CD}$$
induces a map between the long exact sequences of the respective fibrations:
$$\begin{CD}
\dots @>>> \pi_n(F) @>>> \pi_n(E) @>>> \pi_n(B) @>>> \pi_{n-1}(E) @>>> \dots \\
@. @VVV @VVV @VVV @VVV @. \\
\dots @>>> \pi_n(F_p) @>>> \pi_n(E_p) @>>> \pi_n(B) @>>> \pi_{n-1}(E) @>>> \dots
\end{CD}$$
Since $E$ and $E_p$ are homotopy weakly homotopy equivalent ($PB$ is contractible), and so by the five lemma and induction, the maps $\pi_n(F) \to \pi_n(F_p)$ are isomorphisms.
You probably know how the inverse looks for $n=1$: the inverse of a loop $l(t)$ is the same loop but traversed in the opposite direction: $l^{-1}(t) = l(1-t)$. This way, the loop $l l^{-1}$ is homotopic to a constant loop: one can go from $l(0)$ to $l(1)$ and back and this will be $l l^{-1}$, or one can go from $l(0)$ to $l(\tau)$ only and then go back, for some $\tau \in [0,1].$ For $\tau = 1$ gives $l l^{-1}$ and for $\tau = 0$ gives the the loop constantly equal to $l(0)$.
Now, the pinch map interpretation is the same as the composition of loops, except you add some extra dimensions, in which you don't really do anything. Composing two loops is making a circle into the figure "8" and then mapping each of the little circles by each of the loop maps. In higher dimensions, you just perform this operation of every circular section of the sphere. This should give you an idea of how the inverse is supposed to look.
Best Answer
Write $X=S^3\vee S^6$ and note that this space admits at least one comultiplication, since it is a suspension. Denote the suspension comultiplication $$c:X\rightarrow X\vee X$$ and observe that it is coassociative, cocommutative and counital, since $X$ is a double suspension. Thus for each space $Y$, the comultiplication $c$ furnishes the homotopy set $[X,Y]$ with a group structure, which is natural with respect to maps $Y\rightarrow Y'$.
Now recall that the homotopy fibre of the inclusion $j:X\vee X\hookrightarrow X\times X$ is equivalent to $\Sigma \Omega X\wedge \Omega X$ and there is a fibration sequence $$\dots\Omega(X\vee X)\xrightarrow{\Omega j}\Omega X\times \Omega X\xrightarrow{\delta}\Sigma \Omega X\wedge \Omega X\xrightarrow{w} X\vee X\xrightarrow{j} X\times X$$ which defines the maps $w$ and $\delta$.
Consider the resulting Puppe sequence $$\dots\rightarrow [X,\Omega X\times \Omega X]\xrightarrow{\delta_*} [X,\Sigma\Omega X\wedge \Omega X]\xrightarrow{w_*} [X,X\vee X]\xrightarrow{j_*} [X,X\times X],$$ which here is an exact sequence of abelian groups. Since a comultiplication on $X$ is a map $X\rightarrow X\vee X$ which lifts the diagonal $\Delta:X\rightarrow X\times X$ through $j$, all the comultiplications 'live in' $[X,X\vee X]$ and are all mapped down to the same element in $[X,X\times X]$ by $j_*$ .
Thus by exactness we see that the set of comultiplications on $X$ is in bijective correspondence with the image of $w_*$, and so identifies with a certain coset in $[X,\Sigma\Omega X\wedge \Omega X]$. The task now is to identify this coset. Here is the trick: the fibration sequence splits after looping. Indeed, if $pr_i:X\times X\rightarrow X$, $i=1,2$, are the two projections, and $in_i:X\hookrightarrow X\vee X$, $i=1,2$ are the two inclusions, then $$s=\Omega(in_1 pr_1)+\Omega (in_2pr_2):\Omega (X\times X)\cong \Omega X\times \Omega X\rightarrow \Omega (X\vee X)$$ is a section of $\Omega j$. Here I am using the loop addition on $\Omega (X\vee X)$ to form the sum.
Now, since $\Omega j$ admits a section, $\delta$ is null-homotopic: $$\delta\simeq \delta(\Omega w)s\simeq (\delta\Omega w)s\simeq \ast s\simeq \ast.$$ In particular, in the Puppe sequence, $\delta_*=0$, and $w_*:[X,\Sigma\Omega X\wedge \Omega X]\rightarrow[X,X\vee X]$ is monic.
The conclusion is that the coset is the entire group, and fixing one comultiplication induces a bijective correspondence between comultiplications on $X$ and homotopy classes of maps $X\rightarrow \Sigma\Omega X\wedge \Omega X$. All of this follows because of the presence of at least one comultiplication on $X$, namely $c$. The other comultiplications are obtained from $c$ as suitable perturbations. In particular, if $f:X\rightarrow\Sigma \Omega X\wedge \Omega X$ is a map, then $$c'=c+wf$$ is a comultiplication, where the sum is formed using the abelian group structure on $[X,X\vee X]$ which we have fixed with $c$. Of course, if we know already that $X$ admits other comultiplications, we may equally we start with them in place of $c$.
Finally, we come to understand such maps. We have $$[X,\Sigma\Omega X\wedge X]=[S^3\vee S^6,\Sigma\Omega X\wedge \Omega X]\cong \pi_3(\Sigma\Omega X\wedge \Omega X)\oplus\pi_6(\Sigma\Omega X\wedge \Omega X).$$ Using the Hilton-Milnor Theorem we have $$\Omega X=\Omega\Sigma (S^2\vee S^5)\simeq \Omega S^3\times \Omega S^6\times \Omega S^8\times\dots$$ and with a little more work we get $$\Sigma \Omega X\wedge \Omega X\simeq (\Sigma\Omega S^3\wedge \Omega S^3)\vee (\Sigma\Omega S^3\wedge \Omega S^5)\vee (\Sigma\Omega S^3\wedge \Omega S^5)\vee\dots$$ where the omitted terms are at least 8-connected.
Then $\Omega S^3$ is 2-conected and $\Omega S^5$ is 4-connected, so $$\pi_3(\Sigma \Omega X\wedge \Omega X)=0.$$ On the other hand $$\pi_6(\Sigma \Omega X\wedge \Omega X)\cong \pi_6(\Sigma\Omega S^3\wedge \Omega S^3)\cong\pi_6(S^5\cup e^7\cup e^7\cup\dots).$$ A quick calculation in the cohomology on $\Omega S^3\wedge \Omega S^3$ shows that there is no Steenrod square $Sq^2:H^4(\Omega S^3\wedge \Omega S^3;\mathbb{Z}_2)\rightarrow H^6(\Omega S^3\wedge \Omega S^3;\mathbb{Z}_2)$, and since this operations detects the stable class $\eta$, we conclude that $$\Sigma\Omega S^3\wedge \Omega S^3\simeq (S^5\vee S^7\vee S^7)\cup\dots$$ (Note that $H^*\Omega S^3$ is a divided power algebra and is torsion free). In particular $$\pi_6(\Sigma \Omega X\wedge \Omega X)\cong \pi_6(S^5\vee S^7\vee S^7)\cong \pi_6 S^5\cong \mathbb{Z}_2.$$
Putting everything together, comultiplications on $X=S^3\vee S^6$ are in bijective correspondence with the set $$[X,\Sigma\Omega X\wedge \Omega X]\cong\pi_6S^5\cong\mathbb{Z}_2.$$