Let $X$ and $Y$ be some (topological) spaces. The mapping space $X^Y$ is the set of maps $f:X \rightarrow Y$ endowed with the compact-open topology. A map $g:Z \rightarrow Y$ induces a map $g^*:X^Y \rightarrow X^Z$ by pre-composition. Key propery of mapping spaces is the following: Let $i: B\rightarrow Y$ be a cofibration of locally compact Hausdorff spaces, and let $X$ be any space. Then the induced map $i^*:X^Y \rightarrow X^B$ is a fibration. I want to show two things:
- That there is a homeomorphism $X^{\{0,1\}} \cong X \times X$, where $\{0,1\}$ is given the discrete topology.
- That the map $\pi_X:X^I \rightarrow X \times X$ given by $\pi_X(\gamma)=(\gamma(0),\gamma(1))$ is a fibration.
As for the first one, since $\{0,1\}$ is discrete, it seems to me that the only good candidate for a homeomorphism is the map $\pi(\gamma) = (\gamma(0),\gamma(1))$. We could also consider $\pi(\gamma) = (\gamma(0),\gamma(0))$, or $\pi(\gamma) = (\gamma(1),\gamma(1))$ but not these are clearly not surjective, right? Is this correct? I am not sure how to show that $\pi(\gamma)$ is a continuous and open mapping.
As for the second one, I start with the inclusion map $i:\{0,1\} \rightarrow [0,1]$. Since $\{1,0\}$ is a subset of $[0,1]$, $i$ is a cofibration. Moreover, both $\{1,0\}$ and $[0,1]$ are locally compact Hausdorff spaces. The induced map $i^{*}:X^{[0,1]} \rightarrow X^{\{0,1\}}$ is than a fibration. Then I should probably use $1.$ to show that $\pi_X$ is a fibration, but I'm not sure how to get this one. Any help will be much appreciated.
Best Answer
You are not explicit whether "compact" includes Hausdorff, so it is not absolutely clear what the compact-open topology is. Let us assume that the Hausdorff property is not included.
You know that if $g : Z \to Y$ is continuous, then $g^* : X^Y \to X^Z, f \mapsto f \circ g$, is continuous (If "compact" includes Hausdorff, then we must additionally require that $Y$ is Hausdorff to assure that $g^*$ is continuous. In the context of your question this is satisfied.)
Your map $\pi$ is a homeomorphism. Since $\{0,1\}$ is discrete, all functions $f : \{0,1\} \to X$ are continuous which shows that $\pi$ is bijective. A subbasis $\Sigma$ for $X^{\{0,1\}}$ is given by the sets $\langle K, U \rangle = \{ f \in X^{\{0,1\}} \mid f(K) \subset U \}$ where $K \subset \{0,1\}$ is compact and $U \subset X$ is open. Note that all $K \subset \{0,1\}$ are compact. We have $\langle \emptyset, U \rangle = X^{\{0,1\}}$ and $\langle \{0,1\}, U \rangle = \langle \{0\}, U \rangle \cap \langle \{1\}, U \rangle$, thus the set $\Sigma' = \{ \langle \{i\}, U \rangle \mid i = 0,1 , U \subset X \text{ open } \}$ is a subbasis for $X^{\{0,1\}}$. But $\pi(\langle \{0\}, U \rangle = U \times X$ and $\pi(\langle \{1\}, U \rangle = X \times U$, and these sets form a subbasis for the product topology on $X \times X$. Thus $\pi$ is continuous and open, hence a homeomorphism.
The inclusion $i : \{0,1\} \to I$ is a cofibration. The reason is not that $\{0,1\}$ is a subset of $I$, but is true because it is closed subset and $I \times \{0\} \cup \{0,1\} \times I$ is a retract of $I \times I$. You know that $i^* : X^I \to X^{\{0,1\}}$ is a fibration. Hence also $\pi_X = \pi \circ i^*$ is one simply because $\pi$ is a homeomorphism.