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I realize most people work in "convenient categories" where this is not an issue.

In most topology books there is a proof of the fact that there is a natural homeomorphism of function spaces (with the compact-open topology): $$F(X\times Y,Z)\cong F(X,F(Y,Z))$$ when $X$ is Hausdorff and $Y$ is locally compact Hausdorff. There is also supposed to be a homeomorphism in the based case with the same conditions on $X$ and $Y$: $$F_{\ast}(X\wedge Y,Z)\cong F_{\ast}(X,F_{\ast}(Y,Z))$$involving spaces of based maps and the smash product $\wedge$. This, for instance, is asserted on n-lab. I checked the references listed on this page and many other texts but have not found a proof of this "well-known fact."

It seems pretty clear if $X$ and $Y$ are compact Hausdorff (EDIT: in fact this is Theorem 6.2.38 of Maunder's Algebraic Topology) but can this really be proven in this generality?

Can anyone provide a reference for a proof?

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    Can't you get it just from general considerations? The maps in $F_*(X\wedge Y,Z)$ are in one-to-one correspondence with maps in $F(X\times Y,Z)$ that agree on $X\times y_0$ and $x_0\times Y$ and maps them to the base point of $Z$; these maps, under the isomorphism to $F(X,F(Y,Z))$, such an $f$ goes to the map that sends $x$ to $f_x\colon Y\to Z$; since $f$ agrees on all of $X\times{y_0}$, and maps to $z_0$, then $f_x(y_0) = z_0$, so the image is in $F_*(Y,Z)$, and $f_{x_0}$ maps tot he constant function $Y\mapsto z_0$, which is the basepoint of $F_*(Y,Z)$. Isomorphism should follow from lifts.2011-02-04
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    Showing the desired isomorphism is a homeomorphism should require you make use of compact sets in the quotient space $X\wedge Y$. This is non-obvious to me. Are you sure the space $F_{\ast}(X\wedge Y,Z)$ is homeomorphic to the relative mapping space $F((X\times Y,X\vee Y),(Z,z_0))$?2011-02-05

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