From NASA.gov:
“An astronomical object can look very different depending on what wavelength is used in its detection. The object’s appearance often changes across the electromagnetic spectrum because various physical processes result in emission in different wavelength regions. Images of emission can be made by mapping brightness to colors (see “Making an Image”). For example, the optical image of the Andromeda galaxy (below left) shows glowing stars. And dark dust lanes trace out a spiral arm structure. An infrared image of the galaxy (below right) shows several concentric rings of dust rather than spiral arms. The dust is too cold (-260 degree C) to be detected in optical light.
Below are five images of the galaxy M 33. [see link]
Each image is taken in a different wavelength region.
In the X-ray image, we see very hot, diffuse interstellar gas, and bright point-like X-ray sources, such as X-ray binaries.
The brightness in the ultraviolet image indicates star formation activity. Stars are visible in the optical image.
Red supergiant stars and dust heated by massive star formation are highlighted in the near-infrared image, which looks similar to the optical image.
The radio image maps out hydrogen gas in the galaxy. The red coloring is indicative of gas moving away from our line of sight. The blue colors gas moving toward us (a phenomena described by Doppler shift).”
http://mwmw.gsfc.nasa.gov/mmw_across.html
Thank You.
Actually, after thinking about it a little more, if the image we see today is now in the visible range of the EM spectrum (if red-shifted from UV to visible), it will just be the UV information made visible via universal expansion. Normally, UV and the other forms of non-visible EM radiation needs to be converted (color coded) in order to provide an image we can see.