Let $\textbf{Top}$ be the category of all topological spaces, including the bad ones. We can make $\textbf{Top}$ into a simplicially enriched category as follows:
Given topological spaces $X$ and $Y$, define the simplicial set $\textrm{Map} (X, Y)$ by the following formula: $$\textrm{Map} (X, Y)_n = \textrm{Hom}_\textbf{Top} (X \times \left| \Delta^n \right|, Y)$$ The RHS is the set of continuous maps $X \times \left| \Delta^n \right| \to Y$.
Given continuous maps $f : X \times \left| \Delta^n \right| \to Y$ and $g : Y \times \left| \Delta^n \right| \to Z$, define $g \circ_n f : X \times \left| \Delta^n \right| \to Z$ to be the following composite: $$\require{AMScd} \begin{CD} X \times \left| \Delta^n \right| @>{\textrm{id}_X \times \delta_{\left| \Delta^n \right|}}>> X \times \left| \Delta^n \right| \times \left| \Delta^n \right| @>{f \times \textrm{id}_{\left| \Delta^n \right|}}>> Y \times \left| \Delta^n \right| @>{g}>> Z \end{CD}$$ Here, $\delta_{\left| \Delta^n \right|} : \left| \Delta^n \right| \to \left| \Delta^n \right| \times \left| \Delta^n \right|$ is the diagonal embedding.
(Basically we are exploiting the fact that every object – so $\left| \Delta^n \right|$ in particular – in a cartesian monoidal category has a unique comonoid structure. You can do this kind of thing whenever you have a comonoid in a monoidal category.)
This defines a simplicial map $\textrm{Map} (Y, Z) \times \textrm{Map} (X, Y) \to \textrm{Map} (X, Z)$ and it is easy to verify that it is a unital and associative composition.
If $X$ is exponentiable (e.g. locally compact Hausdorff), then we have $$\textrm{Map} (X, Y)_n \cong \textrm{Hom}_\textbf{Top} (\left| \Delta^n \right|, Y^X)$$ so $\textrm{Map} (X, Y) \cong \textrm{Sing} (Y^X)$. The above construction can be regarded as a workaround for $Y^X$ (in the sense of category theory) not always existing.
Since $\left| \Delta^n \right|$ is compact Hausdorff, we always have $$\textrm{Map} (X, Y)_n \cong \textrm{Hom}_\textbf{Top} (X, Y^{\left| \Delta^n \right|})$$ so the above definition is self-dual in some sense.
Question. Is $\textbf{Top}$, with this simplicial enrichment, tensored and cotensored over $\textbf{sSet}$?
There were many subtle points to be checked, but I have convinced myself the answer is yes. Is this in the literature somewhere, or at least, known folklore?
I have seen claims that $\textbf{Top}$ is not a tensored and cotensored simplicially enriched category, which made me doubt my reasoning, but I think there is a subtlety that has been missed. What is true is that there are difficulties making $(X, K) \mapsto X \times \left| K \right|$ into a (right) monoidal action of $\textbf{sSet}$ on $\textbf{Top}$, not least because we do not always have $\left| K \times L \right| \cong \left| K \right| \times \left| L \right|$. Another problem is that $\left| K \right|$ is not always exponentiable. But there is another way!
A straightforward calculation shows that there is only one candidate for the tensor of a topological space $X$ by a simplicial set $K$, given by the following coend formula: $$X \odot K = \int^n X \times \left| \Delta^n \right| \times K_n$$ This is not the same as $X \times \left| K \right|$! At least, it is not obviously the same, but there is a natural continuous bijective map $X \odot K \to X \times \left| K \right|$. Put it another way, $X \odot K$ is $X \times \left| K \right|$ but with a possibly finer topology, namely the finest topology making continuous the maps $X \times \left| \Delta^n \right| \to X \odot K$ corresponding at the point set level to the continuous maps $\textrm{id}_X \times \left| k \right| : X \times \left| \Delta^n \right| \to X \times \left| K \right|$ for each simplicial map $k : \Delta^n \to K$.
Dually, there is only one candidate for the cotensor of $Y$ by $K$, given by the following end formula: $$K \pitchfork Y = \int_n Y^{\left| \Delta^n \right| \times K_n}$$ This is not the same as $Y^{\left| K \right|}$. I do not even claim $\left| K \right|$ is exponentiable! If $Y^{\left| K \right|}$ exists (because e.g. $\left| K \right|$ is exponentiable or $Y$ is indiscrete) then there is a canonical continuous bijective map $Y^{\left| K \right|} \to K \pitchfork Y$. In any case, $K \pitchfork Y$ is the set of continuous maps $\left| K \right| \to Y$ with the coarsest topology making continuous the maps $K \pitchfork Y \to Y^{\left| \Delta^n \right|}$ corresponding at the point set level to precomposition by $\left| k \right| : \left| \Delta^n \right| \to \left| K \right|$ for each simplicial map $k : \Delta^n \to K$.
For $X \odot K$ to have the simplicially enriched universal property of the tensor of $X$ by $K$, it is necessary and sufficient that the following holds:
- A map $f : (X \odot K) \times \left| \Delta^n \right| \to Y$ is continuous if and only if, for every simplicial map $k : \Delta^p \to K \times \Delta^n$, the composite $f \circ (\textrm{id}_X \times \left| k \right|) : X \times \left| \Delta^p \right| \to Y$ is continuous.
Dually, for $K \pitchfork Y$ to have the simplicially enriched universal property of the cotensor of $Y$ by $K$, it is necessary and sufficient that the following holds:
- A map $f : X \to (K \pitchfork Y)^{\left| \Delta^n \right|}$ is continuous if and only if, for every simplicial map $k : \Delta^p \to K \times \Delta^n$, the composite $\left| k \right|^* \circ f : X \to Y^{\left| \Delta^p \right|}$ is continuous, where $\left| k \right|^* : (K \pitchfork Y)^{\left| \Delta^n \right|} \to Y^{\left| \Delta^p \right|}$ is precomposition by $\left| k \right| : \left| \Delta^p \right| \to \left| K \times \Delta^n \right|$.
So in some sense this is all just a matter of very concrete point-set topology, but I only have an indirect abstract proof building on the natural homeomorphism $\left| \Delta^m \times \Delta^n \right| \to \left| \Delta^m \right| \times \left| \Delta^n \right|$.
