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I have been working my way through Lawvere and Schanuel (1997) without too much trouble, but now that I am up to Article V, I am stumped. So, without further ado:

Exercise 6: In a category with products in which map objects exist for any two objects, there is for any three objects a standard map: $B^A \times C^B \stackrel{\gamma}{\longrightarrow} C^A$ which represents composition in the sense that $\gamma\langle\lceil f \rceil, \lceil g \rceil\rangle = \lceil gf \rceil $ for any $A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C.$

I understand how an exponential map object is like a product map, and I can diagram product maps just fine. They're pretty straightfoward. But for this problem, I think the key to solving the problem lies in isolating each of the $\lceil f \rceil, \lceil g \rceil$ terms in order to calculate the product $\lceil gf \rceil$. In the book, they only give them as parameters of the functions f and g taking A to B and B to C, respectively.

When I was first stuck, I looked up the definition of a exponential map object in wikipedia, but that didn't help me in the calculation. I can write the diagram from A to B and from B to C separately, but I am stuck at the point of putting them together into a composite function.

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By universal property we get that there are morphisms $ \cdot_{A,B} \colon B^A \times A \to B$ and $ \cdot_{B,C} \colon C^B \times B \to C$ which are universals.

Then you get the morphism $$ \tilde \gamma \colon C^B \times B^A \times A \stackrel{1_{C^B} \times \cdot_{A,B}}{\longrightarrow} C^B \times B \stackrel{\cdot_{B,C}}{\longrightarrow} C$$ which is the map that let an element of $B^A$ act on $A$ and then let an element of $C^B$ act on the result.

Now consider the corresponding map $$\gamma \colon C^B \times B^A \to C^A$$ this is your composition (up the isomorphism $C^B \times B^A \cong B^A \times C^B$).

I'd rather left the details, but if you need them let me know and I'll try to add them (hoping I'll find the time).

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The level of detail in your answer was perfect! You used the UMP to set up the two maps, brought them together by means of substitution, made a projection from the product to the exponent, and then switched the order, which is correct up to isomorphism (why is that, btw?). –  bwkaplan Jan 28 '12 at 20:26
    
@bwkaplan $C^B \times B^A$ and $B^A \times C^B$ are not necessarily the same object, some times are simply isomorphic, the arrow that I found is of type $C^B \times B^A \to C^A$ while you were looking for a morphism of type $B^A \times C^B \to C^A$, so in order to to obtain it you have to compose the morphism I've found with the isomorphism $B^A \times C^B \cong C^B \times B^A$. –  Giorgio Mossa Jan 29 '12 at 16:39
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