j\bigr\}\subset K^{c_{{\delta},1}}.$$

On the other hand, take $R$ large enough such that $\inf\limits_{\Omega}q\ge
-\sum_{i=1}^m\frac{(p-1)\kappa_i^2}{16m\pi\min_l\{\kappa_l\}}-\sum_{i=1}^m\frac{\kappa_i^2g(z_j,z_j)}{4m\pi\min_l\{\kappa_l\}}$, then $K(Z)\le c_{{\delta},1}$ implies that

$$-\sum_{j\ne i}^m \frac{\pi{\delta}^2\kappa_i\kappa_j \bar
G(z_j,z_i)}{(\ln\frac{R}{{\varepsilon}})^2}\le
-\frac{\pi{\delta}^2\ln\frac1{\varrho}}{|\ln{\varepsilon}||\ln\frac{R}{{\varepsilon}}|},$$ which implies that there are $i\ne j$, such that

$$\bar G(z_j,z_i)\ge c'\ln\frac1{\varrho},\quad\text{where}~~c'>0
~~\text{is a constant}.$$ So, there is a $\alpha>0$, independent of $\delta$, such that

$$|z_i-z_j|\le \varrho^\alpha.$$ Therefore,

$$\label{4.6}
\begin{split}
& \bigl\{Z:  Z\in D,  |z_i-z_j|=\varrho^{\bar L}, \;\text{for
some}\; i\ne
j\bigr\}\\
&\subset K^{c_{{\delta},1}}\subset \bigl\{Z:  Z\in D, |z_i-z_j|\le
\varrho^\alpha, \;\text{for some}\; i\ne j\bigr\}.
\end{split}$$

Filling the hole $D^*=: \bigl\{Z:  Z\in D,  |z_i-z_j|=\varrho^{\bar
L},
 \;\text{for some}\; i\ne
j\bigr\}$ in $D$, we obtain

$$\label{4.7}
\begin{split}
& \bigl\{Z:  z_i\in\Omega_\varrho,
 |z_i-z_j|\le \varrho^{\bar L}, \;\text{for some}\; i\ne
j\bigr\}\\
&\subset K^{c_{{\delta},1}}\cup D^* \subset \bigl\{Z:
z_i\in\Omega_\varrho,
 |z_i-z_j|\le
\varrho^\alpha, \;\text{for some}\; i\ne j\bigr\}.
\end{split}$$

Since $K^{c_{{\delta},1}}$ is a deformation retract of $K^{c_{{\delta},2}}$, we find that $K^{c_{{\delta},1}}\cup D^*$ is a deformation retract of $K^{c_{{\delta},2}} \cup D^*$. On the other hand, $\bigl\{Z:  z_i\in\Omega_\varrho,  z_i=z_j, \;\text{for some}\; i\ne
j\bigr\}$ is a deformation retract of $\bigl\{Z:
z_i\in\Omega_\varrho,
 |z_i-z_j|\le
\varrho^\alpha, \;\text{for some}\; i\ne j\bigr\}$ if $\varrho>0$ is small. Using , we see that

$$\bigl\{Z:  z_i\in\Omega_\varrho,  z_i=z_j, \;\text{for some}\; i\ne
j\bigr\}$$ is a deformation retract of $$\underbrace{\Omega_\varrho\times\cdots\times \Omega_\varrho}_m=
K^{c_{{\delta},2}} \cup D^*.$$ This is impossible if $\Omega$ has nontrivial homology.

Thus we 