View source for Bifunctor

A mapping <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161901.png" />, defined on the Cartesian product of two categories <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161902.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161903.png" /> with values in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161904.png" />, which assigns to each pair of objects <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161905.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161906.png" /> some object <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161907.png" />, and to each pair of morphisms

<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161908.png" /></td> </tr></table>

the morphism

<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b0161909.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>

The following conditions

<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b01619010.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2)</td></tr></table>

<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b01619011.png" /></td> </tr></table>

must also be met. In such a case one says that the functor <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b01619012.png" /> is contravariant with respect to the first argument and covariant with respect to the second.



====Comments====
What is described above is a bifunctor contravariant in its first argument and covariant in its second. A bifunctor covariant in both arguments, the more fundamental notion ([[#References|[a1]]]), has (1) and (2) replaced by

<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b01619013.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1prm)</td></tr></table>

<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b016/b016190/b01619014.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2prm)</td></tr></table>

Similarly one can define bifunctors contravariant in both arguments and covariant in the first and contravariant in the second argument.

====References====
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  B. Mitchell,   "Theory of categories" , Acad. Press  (1965)</TD></TR></table>