# Applications of the exponential law $Z^{X \times Y} \approx (Z^Y)^X$

Let $$B^A$$ denote the space of continuous maps $$A \to B$$ with the compact-open topology. I shall not be specific in the interpretation of compact (i.e. whether it includes Hausdorff or not). You can take your favorite interpretation.

It is well-known that for locally compact $$Y$$, there is a canonical bijection $$E : Z^{X \times Y} \to (Z^Y)^X .$$ This has a plethora of applications in topology.

So let $$Y$$ be locally compact. Then it is known that the bijection $$E$$ is even a homeomorphism under suitable additional assumptions (for example $$X, Y$$ Hausdorff).

This is of course an interesting theorem, but does the fact that $$E$$ is a homeomorphism have any applications in general or algebraic topology? I am not aware of any.

• Should you not have your assumptions on $Y$ rather than $X$? (since what you really need for the map to be bijective is that the evaluation $Z^Y\times Y\rightarrow Z$ is continuous). – Tyrone Mar 14 '20 at 13:02
• @Tyrone The assumption on $Y$ is local compactness. However, this is not sufficient and we need additional assumptions on $X,Y$. If we understand that compact and locally compact include Hausdorff, then we need $X$ Hausdorff. – Paul Frost Mar 14 '20 at 13:12
• @Tyrone I realized that I had a typo ("It is well-known that for locally compact $X$ ..."). This certainly caused confusion. – Paul Frost Mar 14 '20 at 13:34
• Now I am happy :D (I was only addressing the typo you just fixed before you wrote down the map $E$). Here is my application: if $\psi:f\simeq g:X\rightarrow Y$ is a homotopy, then $Z^\psi:Z^Y\rightarrow Z^{X\times I}\cong (Z^X)^I$ is a homotopy of the induced maps $Z^f, Z^g$. – Tyrone Mar 14 '20 at 13:35
• @Tyrone Perhaps you should give an official answer. By the way, it would be sufficient to know that $E$ is continuous. – Paul Frost Mar 14 '20 at 13:45

Here is a useful application in homotopy theory. It turns out that understanding the homotopy groups of function spaces is greatly facilitated by the use of adjunction maps. This is already very useful, but when we can arrange that these homotopy groups characterise the space entirely, then we can say some really powerful things.

Let $$A$$ be your favourite finitely generated abelian group and form the Eilenberg-Mac Lane space $$K(A,n)$$. Let $$X$$ be a finite, connected CW complex and consider the function space $$Map(X,K(A,n))=K(A,n)^X.$$ It is a theorem of R. Thom that this function space is itself a generalised Eilenberg-Mac Lane space (i.e. a product of Eilenberg-Mac Lane spaces in various dimensions). Here are the details.

To begin we apply a Theorem of J. Milnor, which says that if $$X,Y$$ are CW complexs and $$X$$ is finite, then the function space $$Map(X,Y)$$ is homotopy equivalent to a CW complex. Thus under the above assumptions, $$Map(X,K(A,n))$$ has CW homotopy type, so to show that it a GEM it will be sufficient to (carefully) check its homotopy groups.

To this end we make use of the fact that CW complexes Hausdorff and finite CW complexes are locally compact. This means that our adjunction maps $$E$$ are homeomorphisms, and in particular are homotopy equivalences. Using them we have $$\pi_kMap(X,K(A,n))\cong\pi_0\left(Map_*(S^k,Map(X,K(A,n))\right)\cong Map_*(S^k\rtimes X,K(A,n)))\cong \pi_0(Map(X,\Omega^kK(A,n)).$$ Here $$Map_*$$ denotes the space of pointed maps, $$S^k\rtimes X=(S^k\times X)/\ast\times X$$ is the half-smash product which we need to introduce to make sense of the adjunctions, and $$\Omega^kK(A,n)=Map_*(S^k,K(A,n))$$ is the $$k$$-fold loop space of $$K(A,n)$$.

Now $$\Omega^kK(A,n)\simeq K(A,n-k)$$ (this is a point if $$k>n$$), so the last line in the equation above is just the set of unbased homotopy classes $$X\rightarrow K(A,n-k)$$. But here we have $$[X,K(A,n-k)]\cong H^{n-k}(X;A).$$ In particular $$\pi_k\left(Map(X,K(A,n))\right)\cong H^{n-k}(X;A).$$

Now the evaluation $$ev:Map(X,K(A,n))\times X\rightarrow K(A,n)$$ gives a cohomology class in $$H^n(Map(X,K(A,n))\times X;A)$$, and taking slant products with $$H_0(X;H^{n-k}X)\cong H^{n-k}(X)$$ we get cohomology classes in $$H^n(Map(X,K(A,n));H^{n-k}(X;A))$$ (use universal coefficients to identify the coefficient group). These classes are exactly homotopy classes of maps $$Map(X,K(A,n))\rightarrow K(H^{n-k}(X;A),k)$$. Thom shows that the collection of these maps assemble into a homotopy equivalence $$Map(X,K(A,n))\simeq\prod_{k=0}^nK(H^{n-k}(X,A),k)$$ giving us our claim.

• +1 for your answer. I am waiting whether there will be other answers before I acknowldege it. – Paul Frost Mar 15 '20 at 22:59