create a canvas representation of your PNG. The following creates an off-screen canvas that is the same width and height as your image and has the image drawn on it.
var img = document.getElementById('my-image');
var canvas = document.createElement('canvas');
canvas.width = img.width;
canvas.height = img.height;
canvas.getContext('2d').drawImage(img, 0, 0, img.width, img.height);
After that, when a user clicks, use event.offsetX and event.offsetY to get the position. This can then be used to acquire the pixel:
var pixelData = canvas.getContext('2d').getImageData(event.offsetX, event.offsetY, 1, 1).data;
Building on Jeff's answer, your first step would be to create a canvas representation of your PNG. The following creates an off-screen canvas that is the same width and height as your image and has the image drawn on it.
var img = document.getElementById('my-image');
var canvas = document.createElement('canvas');
canvas.width = img.width;
canvas.height = img.height;
canvas.getContext('2d').drawImage(img, 0, 0, img.width, img.height);
After that, when a user clicks, use `event.offsetX` and `event.offsetY` to get the position. This can then be used to acquire the pixel:
var pixelData = canvas.getContext('2d').getImageData(event.offsetX, event.offsetY, 1, 1).data;
Because you are only grabbing one pixel, pixelData is a four entry array containing the pixel's R, G, B, and A values. For alpha, anything less than 255 represents some level of transparency with 0 being fully transparent.
Here is a jsFiddle example: http://jsfiddle.net/thirtydot/9SEMf/869/ I used jQuery for convenience in all of this, but it is by no means required.
**Note:** `getImageData` falls under the browser's same-origin policy to prevent data leaks, meaning this technique will fail if you dirty the canvas with an image from another domain or (I believe, but some browsers may have solved this) SVG from any domain. This protects against cases where a site serves up a custom image asset for a logged in user and an attacker wants to read the image to get information. You can solve the problem by either serving the image from the same server or implementing [Cross-origin resource sharing](http://en.wikipedia.org/wiki/Cross-origin_resource_sharing).
