Drupal - Snippet to programatically return a view as a response to AJAX request from controller

myController.php

<?php

use Drupal\Core\Ajax\AjaxResponse;
use Drupal\views\Views; 
use Drupal\Core\Render\Renderer

/**
   * Drupal\Core\Render\Renderer definition.
   *
   * @var \Drupal\Core\Render\Renderer
   */
  protected $renderer;

  /**
   * Drupal\Core\Form\FormBuilder definition.
   *
   * @var \Drupal\Core\Form\FormBuilder
   */
  protected $formBuilder;

/**
   * constructor.
   */
  public function __construct(Renderer $renderer, FormBuilder $formBuilder, BlockManager $pluginManagerBlock) {
    $this->renderer = $renderer;
    $this->formBuilder = $formBuilder;
  }

  /**
   * {@inheritdoc}
   */
  public static function create(ContainerInterface $container) {
    return new static(
      $container->get('renderer'),
      $container->get('form_builder'),
    );
  }

$view = Views::getView($view);     
if ($view) { 
  $view->setDisplay($display); 
  $view->initHandlers();
  $form_state = (new FormState()) 
  ->setStorage([ 
  'view' => $view, 
  'display' => &$view->display_handler->display, 
  'rerender' => TRUE, 
  ]) 
  ->setMethod('get') 
  ->setAlwaysProcess() 
  ->disableRedirect(); 
  $form_state->set('rerender', NULL); 
  $html = $this->formBuilder->buildForm('\Drupal\views\Form\ViewsExposedForm', $form_state); 
}
$ajax_response = new AjaxResponse();
$target = "#your_id" // or ".classname" of HTML
$ajax_response->addCommand(new HtmlCommand($target, $html));
return $ajax_response;
?>