Assume I have simplicial sets $X:\Delta^{op}\rightarrow Set$ and $Y:\Delta^{op}\rightarrow Set$, then I can form the simplicial set $\operatorname{Hom}(X,Y)_n := \operatorname{Hom}_{sSet}(\Delta^n\times X,Y)$ where $\Delta^n := \operatorname{Hom}_{\Delta}(\cdot,[n])$. Then I have the 'adjunction relation':

$\operatorname{Hom}(Z,\operatorname{Hom}(X,Y)) \simeq \operatorname{Hom}(Z\times X,Y)$

Can I do the same thing for simplicial topological spaces $X,Y:\Delta^{op}\rightarrow \operatorname{Top}$? Are there references on that or is that trivially the same thing? What would be $\Delta^n$ and what topologies do you take on the $\operatorname{Hom}(X,Y)_n$? In particular, I would like to have the same adjunction relation: $\operatorname{Hom}(Z,\operatorname{Hom}(X,Y)) \simeq \operatorname{Hom}(Z\times X,Y)$ but I can't figure out the details of how to construct the simplicial topological space $\operatorname{Hom}(X,Y)$.

  • 1
    $\begingroup$ You could make it into a simplicial topological space if $X_n$ and $Y_n$ are sufficiently nice, but you can definitely make it into a simplicial set, using this end formula: $$\textrm{Hom}(X, Y)_n = \int_{[k] : \mathbf{\Delta}} \textbf{Set}(\mathbf{\Delta}([k], [n]), \textbf{Top}(X_k, Y_k))$$ $\endgroup$ – Zhen Lin Feb 15 '13 at 17:59

Here is some general nonsense that will do what you want. Suppose $\mathcal{C}$ is a locally small cartesian-closed category with limits and colimits for all small diagrams (so, for example, $\textbf{CGHaus}$ or $\textbf{Set}$) and suppose $\mathcal{D}$ is a small category. Then the functor category $[\mathcal{D}, \mathcal{C}]$ will be a cartesian-closed category with limits and colimits for all small diagrams as well. The second part is a standard fact, so I will only explain why $[\mathcal{D}, \mathcal{C}]$ is cartesian-closed.

Let $Y$ and $Z$ be objects in $[\mathcal{D}, \mathcal{C}]$, i.e. functors $\mathcal{D} \to \mathcal{C}$. We define a new object $Z^Y$ using the following formula: $$Z^Y (t) = \int_{d : \mathcal{D}} {Z (d)}^{\mathcal{D}(t, d) \otimes Y (d)}$$ The integral notation refers to ends, and $\otimes$ is defined by the adjunction below: $$\mathcal{C}(A \otimes C', C) \cong \textbf{Set}(A, \mathcal{C}(C', C))$$ The set of natural transformations can also be computed by an end: $$[\mathcal{D}, \mathcal{C}](X, W) \cong \int_{t : \mathcal{D}} \mathcal{C} (X (t), W (t))$$ Moreover, $\mathcal{C}(C, -)$ preserves ends, so we have $$[\mathcal{D}, \mathcal{C}] \left( X, Z^Y \right) \cong \int_{t : \mathcal{D}} \int_{d : \mathcal{D}} \mathcal{C} \left( X (t), {Z (d)}^{\mathcal{D}(t, d) \otimes Y (d)} \right)$$ and by the definition of exponential objects in $\mathcal{C}$, $$[\mathcal{D}, \mathcal{C}] \left( X, Z^Y \right) \cong \int_{t : \mathcal{D}} \int_{d : \mathcal{D}} \mathcal{C} \left( X (t) \times (\mathcal{D}(t, d) \otimes Y (d)), Z (d) \right)$$ and since $C \times (-) : \mathcal{C} \to \mathcal{C}$ has a right adjoint, it preserves $\otimes$, i.e. $$X (t) \times (\mathcal{D}(t, d) \otimes Y (d)) \cong \mathcal{D}(d, t) \otimes (X (t) \times Y (d))$$ and so, using the definition of $\otimes$, we obtain \begin{align} [\mathcal{D}, \mathcal{C}] \left( X, Z^Y \right) & \cong \int_{t : \mathcal{D}} \int_{d : \mathcal{D}} \textbf{Set} \left( \mathcal{D}(t, d), \mathcal{C} \left( X (t) \times Y (d), Z (d) \right) \right) \\ & \cong \int_{d : \mathcal{D}} \int_{t : \mathcal{D}} \textbf{Set} \left( \mathcal{D}(t, d), \mathcal{C} \left( X (t) \times Y (d), Z (d) \right) \right) \end{align} where the second isomorphism is the interchange theorem for ends; but the end version of Yoneda lemma says $$\int_{t : \mathcal{D}} \textbf{Set} \left( \mathcal{D}(t, d), F t \right) \cong F d$$ for any functor $F : \mathcal{D}^\textrm{op} \to \textbf{Set}$, hence, $$[\mathcal{D}, \mathcal{C}] \left( X, Z^Y \right) \cong \int_{d : \mathcal{D}} \mathcal{C} \left( X (d) \times Y (d), Z (d) \right) \cong [\mathcal{D}, \mathcal{C}] (X \times Y, Z)$$ and this isomorphism is natural in $X$, $Y$, and $Z$.

  • $\begingroup$ Thanks! That should do it. $\endgroup$ – Michael Scarn Feb 17 '13 at 17:16
  • $\begingroup$ Ok, I'm still a bit confused about your definition of $\otimes$. It seems to be the right adjoint of the hom-set-functor? Does it exist in this general case you describe? Can you give it explicitly? $\endgroup$ – Michael Scarn Feb 20 '13 at 20:50
  • $\begingroup$ My definition just a functorial way of saying, take a coproduct of such-and-such-many copies of so-and-so. (It is a left adjoint, by the way.) $\endgroup$ – Zhen Lin Feb 20 '13 at 21:54
  • $\begingroup$ Ahh..so you use the set A is an 'index set'. Then I get it. $\endgroup$ – Michael Scarn Feb 20 '13 at 22:43

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