The second statement follows immediately from the construction of $i_n$, which forces $f' ( \beta,\gamma) = $f( \beta$f' ( \beta,\gamma) = f( \beta , \gamma \circ i_n) = \alpha(x)$ which is exactly , \gamma \circ i_n) = \alpha(x)$ which is exactly $f \circ k_n$$f \circ k_n$.
The second statement follows immediately from the construction of $i_n$, which forces $f' ( \beta,\gamma) = $f( \beta , \gamma \circ i_n) = \alpha(x)$ which is exactly $f \circ k_n$.
The second statement follows immediately from the construction of $i_n$, which forces $f' ( \beta,\gamma) = f( \beta , \gamma \circ i_n) = \alpha(x)$ which is exactly $f \circ k_n$.
The significance of the spaces $Y_n$ and $Z_n$ is that $Z_n$ is the set of continuous $I$-valued functions of $x \in X$ that depend only on $e(t,x), e(x,x), e(x,t)$$ e(x,x), e(x,t)$ for $t \in X_n$. $Y_n$ is the space of possible values of the triplepair $e(t,x), e(x,x), e(x,t)$$e(x,x), e(x,t)$, so that $Z_n$ is the space of functions on $Y_n$. Then $X_{n+1}$ is constructed so that it maps surjectively to $Z_n$, so all functions of $Z_n$ come from elements of $X_n$, and maps to $Y_n$, so that all functions of $Z_n$ can be extended to functions on $X_n$. For simplicity and canonicality, we define $X_{n+1}$ to be a subset of $Y_n \times Z_n$, defined by coherence condition to ensure the desired relationship between $Y_n$ and $e$. Then we can freely add also the functions $e(t,x)$ for $t\in X_n$ to $Z_n$, as they are already continuous functions of $e(x,x)$, $e(x,t)$ by construction of $X_n$.
Let $Y_n$ be the subset of $I^{X_n} \times I \times I^{X_n}$ consisting of tuples (a,b,c) satisfying the coherence conditions:
$(a \circ i_{n-1} , b, c \circ i_{n-1}) \in Y_{n-1}$
For all $x$ in $X_{n}$ which corespond to a tuple $(a',b',c'),f'$ in $Y_{n-1} \times Z_{n-1}$, $a(x)= f'( a \circ i_{n-1} ,b, c \circ i_{n-1})$.
In the case $n=0$, so $X_n = \emptyset$, this is vacuous, and so we need not introduce $Y_{-1}$ and $Z_{-1}$$ I \times I^{X_n}$.
Let $Z_n= I^{Y_n}$.
Let $X_{n+1} $ be the subset of $Y_n \times Z_n$ consisting of tuples $(a,b,c)\in Y_n, f\in Z_n$$(b,c)\in Y_n, f\in Z_n$ satisfying the following two coherence conditions:
$b= f(a,b,c)$$b= f(b,c)$
For all $x$ in $X_n$, $c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$$c(x) = f( e_n(x,x), t \mapsto e_n(x,t))$
Let $i_n: X_n \to X_{n+1}$ send $x$$x= ((b',c'),f)$ to $( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t) ) , f(\alpha,\beta,\gamma)=\alpha(x)$$(e_n(x,x), t \mapsto e_n(x,t) ) , (\beta,\gamma)\mapsto f'(\beta, \gamma \circ i_{n-1} )$.
Let $e_{n+1}: X_{n+1} \times X_{n+1} \to I$ send $((a_1,b_1,c_1),f_1),((a_2,b_2,c_2),f_2)$$((b_1,c_1),f_1),((b_2,c_2),f_2)$ to $f_1(a_2,b_2,c_2)$$f_1(b_2,c_2)$.
Let us first check that for $x$ in $X_n$ corresponding to a tuple $(a,b,c),f$$(b,c),f$ in $Y_{n-1} \times Z_{n-1}$, $i_n(x)$ satisfies the coherence conditions.
For $x'$$x$ in $X_n$ which corresponds to a tuple $(a',b',c'),f'$$(b,c),f$, we need $e_n(x',x) = f'(t \mapsto e_n( i_{n-1}(t),x), e_n(x,x), t \mapsto e_n(x,i_{n-1}(t))$. So$e_n(x,x) = f( e_n(x,x), t \mapsto e_n(x, i_{n-1}(t))$, but $e_n(x)=f(b,c)$ so it is sufficient to check that $a(t) = e_n( i_{n-1}(t),x)$, $b=e_n(x,x)$, and $c=e_n(x,i_{n-1}(t))$$c= t \mapsto e_n(x, i_{n-1}(t))$, which follow fromare the definitioncoherence conditions of $e_n$ and respectively the$X_n$.
Furthermore we need for $x'$ in $X_n$ corresponding to a tuple $(b',c'),f)$, $e_n(x,x') = f( e_n(x',x'),t \mapsto e_n(x',i_{n-1}(t)))$, which is true because by definition of $i_{n-1}$$e_n$, the coherence condition for $b$,$e_n(x,x') = f(b',c')$ and the$b' = f'(b',c') =e_n(x',x')$ while $c'(t) = f'( e_{n-1}(t,t), s \mapsto e_{n-1}(s,t)) = e_n(x', i_{n-1}(t))$ by definition of $i_{n-1}$ as well as the coherence condition for$e_n$ and $c$$i_{n-1}$.
The fact that $e_{n+1}(i_n(x_1),i_n(x_2))=e_n(x_1,x_2)$ follows from unwinding the definitions.
Indeed, let $x_1 = ((b_1,c_1),f_1)$ and let $f_2=((b_2,c_2),f_2))$ and then $e_{n+1} (i_n ( x_1),i_n(x_2)) = f_1 ( e_n (x_2,x_2), t \mapsto e_n(x_2, i_{n-1}(t)) =f_1(b_2,c_2)= e_n(x_1,x_2)$ by the coherence conditions for $b_2,c_2$ and the definition of $e_n$, and then the definition of $e_n$.
Consider the map $j_n: Y_{n+1} \to Y_{n}$ that sends $(a,b,c)$$(b,c)$ to $(a \circ i_n, b, c \circ i_n)$$( b, c \circ i_n)$. Consider also the map $k_n: Z_{n} \to Z_{n+1}$ by exponentiating $j_n$.
I claim that if $i_{n+1} ((a,b,c),f)= ((a',b',c'),f')$$i_{n} ((b,c),f)= ((b',c'),f')$ then $j_n(a',b',c') = (a,b,c)$$j_n(b',c') = (b,c)$ and $k_n(f)=f'$.
The first statement is simply the fact that $ a(t)= t \mapsto e_n(i_{n-1}(t),x), b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$$ b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$, the first of which follows from the definition of $e_n$ and the second and thirdfollow from the definition of $e_n$ and the coherence conditions for $b$ and $c$ respectively.
The second statement is simplyfollows immediately from the statement that for $ (\alpha,\beta,\gamma)$ inconstruction of $Y_{n+1}$$i_n$, $f( \alpha \circ i_n, \beta , \gamma \circ i_n) = \alpha(x)$ which is the coherence condition forforces $a$ applied to$f' ( \beta,\gamma) = $f( \beta $\alpha$, \gamma \circ i_n) = \alpha(x)$ which is exactly $f \circ k_n$.
To do this, it is sufficient to prove that the projection map $X_n \to Z_n$ is surjective. In other words, given $f: Y_n \to I$, construct $(a,b,c) \in Y_n$$(b,c) \in Y_n$ satisfying the coherence conditions. Clearly we must have $c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$. For now, let $b$ be a parameter. Define $c_m: X_m \to I$ for $m$ from $0$ to $n$ such that $c_n= c$, $c_m = c_{m+1} \circ i_m$, then define $a_m: X_m \to I$ so that $a_0$ is the empty function and $a_m(x)$ for $x \in X_m$ corresponding to $(a',b',c',f')$ is $f'(a_{m-1}, b, c_{m-1})$$c(x) = f( e_n(x,x), t \mapsto e_n(x,t))$. Then by the definition of $i_{m-1}$, $a_{m-1} =a_m \circ i_{m-1}$. Now each $a_m$ is a continuous function of the previous, so $a_n$ is a continuous function of $b$, hence $b\mapsto f(a_n,b,c)$ has a fixed point. So we maycan take $a=a_n$, $b$ to be theany fixed point andof $c=c$$f(b,c)$.
What maps can we construct this way? We get all the maps defined by elements of $Z_n$ for all $n$. These, in turn, are all continuous maps that depend only on the data in $Y_n$. By the coherence conditions and the definition of $e$, the projection of $x \in X_n$ to $Y_n$ encode $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$$e(x,x),t\mapsto e(x,t)$ for $t$ in $X_{n-1}$. The constructionUsing the definition of $e$ and the projectioncompatibility of $j_n: Y_{n+1} to Y_n$ ensure$e$ with $i_{n-1}$, one can see that it also encodes $t \mapsto e(t,x)$ for $t \in Y_{n-1}$. The compatibility of $i_n$ with $e$ and with $j_n$ ensure that $Y_n$ continues to encode this data for all $x$. Hence all continuous functions that depend only on $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$, which seem to me to be all the "obvious" ones that we construct, are in the image of the constructed map $X \to I^X$.
The significance of the spaces $Y_n$ and $Z_n$ is that $Z_n$ is the set of continuous $I$-valued functions of $x \in X$ that depend only on $e(t,x), e(x,x), e(x,t)$ for $t \in X_n$. $Y_n$ is the space of possible values of the triple $e(t,x), e(x,x), e(x,t)$, so that $Z_n$ is the space of functions on $Y_n$. Then $X_{n+1}$ is constructed so that it maps surjectively to $Z_n$, so all functions of $Z_n$ come from elements of $X_n$, and maps to $Y_n$, so that all functions of $Z_n$ can be extended to functions on $X_n$. For simplicity and canonicality, we define $X_{n+1}$ to be a subset of $Y_n \times Z_n$, defined by coherence condition to ensure the desired relationship between $Y_n$ and $e$.
Let $Y_n$ be the subset of $I^{X_n} \times I \times I^{X_n}$ consisting of tuples (a,b,c) satisfying the coherence conditions:
$(a \circ i_{n-1} , b, c \circ i_{n-1}) \in Y_{n-1}$
For all $x$ in $X_{n}$ which corespond to a tuple $(a',b',c'),f'$ in $Y_{n-1} \times Z_{n-1}$, $a(x)= f'( a \circ i_{n-1} ,b, c \circ i_{n-1})$.
In the case $n=0$, so $X_n = \emptyset$, this is vacuous, and so we need not introduce $Y_{-1}$ and $Z_{-1}$.
Let $Z_n= I^{Y_n}$.
Let $X_{n+1} $ be the subset of $Y_n \times Z_n$ consisting of tuples $(a,b,c)\in Y_n, f\in Z_n$ satisfying the following two coherence conditions:
$b= f(a,b,c)$
For all $x$ in $X_n$, $c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$
Let $i_n: X_n \to X_{n+1}$ send $x$ to $( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t) ) , f(\alpha,\beta,\gamma)=\alpha(x)$.
Let $e_{n+1}: X_{n+1} \times X_{n+1} \to I$ send $((a_1,b_1,c_1),f_1),((a_2,b_2,c_2),f_2)$ to $f_1(a_2,b_2,c_2)$.
Let us first check that for $x$ in $X_n$ corresponding to a tuple $(a,b,c),f$ in $Y_{n-1} \times Z_{n-1}$, $i_n(x)$ satisfies the coherence conditions.
For $x'$ in $X_n$ which corresponds to a tuple $(a',b',c'),f'$, we need $e_n(x',x) = f'(t \mapsto e_n( i_{n-1}(t),x), e_n(x,x), t \mapsto e_n(x,i_{n-1}(t))$. So it is sufficient to check that $a(t) = e_n( i_{n-1}(t),x)$, $b=e_n(x,x)$, and $c=e_n(x,i_{n-1}(t))$, which follow from the definition of $e_n$ and respectively the definition of $i_{n-1}$, the coherence condition for $b$, and the definition of $i_{n-1}$ as well as the coherence condition for $c$.
The fact that $e_{n+1}(i_n(x_1),i_n(x_2))=e_n(x_1,x_2)$ follows from unwinding the definitions.
Consider the map $j_n: Y_{n+1} \to Y_{n}$ that sends $(a,b,c)$ to $(a \circ i_n, b, c \circ i_n)$. Consider also the map $k_n: Z_{n} \to Z_{n+1}$ by exponentiating $j_n$.
I claim that if $i_{n+1} ((a,b,c),f)= ((a',b',c'),f')$ then $j_n(a',b',c') = (a,b,c)$ and $k_n(f)=f'$.
The first statement is simply the fact that $ a(t)= t \mapsto e_n(i_{n-1}(t),x), b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$, the first of which follows from the definition of $e_n$ and the second and third from the definition of $e_n$ and the coherence conditions for $b$ and $c$ respectively.
The second statement is simply the statement that for $ (\alpha,\beta,\gamma)$ in $Y_{n+1}$, $f( \alpha \circ i_n, \beta , \gamma \circ i_n) = \alpha(x)$ which is the coherence condition for $a$ applied to $\alpha$.
To do this, it is sufficient to prove that the projection map $X_n \to Z_n$ is surjective. In other words, given $f: Y_n \to I$, construct $(a,b,c) \in Y_n$ satisfying the coherence conditions. Clearly we must have $c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$. For now, let $b$ be a parameter. Define $c_m: X_m \to I$ for $m$ from $0$ to $n$ such that $c_n= c$, $c_m = c_{m+1} \circ i_m$, then define $a_m: X_m \to I$ so that $a_0$ is the empty function and $a_m(x)$ for $x \in X_m$ corresponding to $(a',b',c',f')$ is $f'(a_{m-1}, b, c_{m-1})$. Then by the definition of $i_{m-1}$, $a_{m-1} =a_m \circ i_{m-1}$. Now each $a_m$ is a continuous function of the previous, so $a_n$ is a continuous function of $b$, hence $b\mapsto f(a_n,b,c)$ has a fixed point. So we may take $a=a_n$, $b$ to be the fixed point and $c=c$.
What maps can we construct this way? We get all the maps defined by elements of $Z_n$ for all $n$. These, in turn, are all continuous maps that depend only on the data in $Y_n$. By the coherence conditions and the definition of $e$, the projection of $x \in X_n$ to $Y_n$ encode $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$ for $t$ in $X_{n-1}$. The construction of the projection $j_n: Y_{n+1} to Y_n$ ensure that it continues to encode this data for all $x$. Hence all continuous functions that depend only on $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$, which seem to me to be all the "obvious" ones that we construct, are in the image of the constructed map $X \to I^X$.
The significance of the spaces $Y_n$ and $Z_n$ is that $Z_n$ is the set of continuous $I$-valued functions of $x \in X$ that depend only on $ e(x,x), e(x,t)$ for $t \in X_n$. $Y_n$ is the space of possible values of the pair $e(x,x), e(x,t)$, so that $Z_n$ is the space of functions on $Y_n$. Then $X_{n+1}$ is constructed so that it maps surjectively to $Z_n$, so all functions of $Z_n$ come from elements of $X_n$, and maps to $Y_n$, so that all functions of $Z_n$ can be extended to functions on $X_n$. For simplicity and canonicality, we define $X_{n+1}$ to be a subset of $Y_n \times Z_n$, defined by coherence condition to ensure the desired relationship between $Y_n$ and $e$. Then we can freely add also the functions $e(t,x)$ for $t\in X_n$ to $Z_n$, as they are already continuous functions of $e(x,x)$, $e(x,t)$ by construction of $X_n$.
Let $Y_n$ be $ I \times I^{X_n}$. Let $Z_n= I^{Y_n}$.
Let $X_{n+1} $ be the subset of $Y_n \times Z_n$ consisting of tuples $(b,c)\in Y_n, f\in Z_n$ satisfying the following two coherence conditions:
$b= f(b,c)$
For all $x$ in $X_n$, $c(x) = f( e_n(x,x), t \mapsto e_n(x,t))$
Let $i_n: X_n \to X_{n+1}$ send $x= ((b',c'),f)$ to $(e_n(x,x), t \mapsto e_n(x,t) ) , (\beta,\gamma)\mapsto f'(\beta, \gamma \circ i_{n-1} )$.
Let $e_{n+1}: X_{n+1} \times X_{n+1} \to I$ send $((b_1,c_1),f_1),((b_2,c_2),f_2)$ to $f_1(b_2,c_2)$.
Let us first check that for $x$ in $X_n$ corresponding to a tuple $(b,c),f$ in $Y_{n-1} \times Z_{n-1}$, $i_n(x)$ satisfies the coherence conditions.
For $x$ in $X_n$ which corresponds to a tuple $(b,c),f$, we need $e_n(x,x) = f( e_n(x,x), t \mapsto e_n(x, i_{n-1}(t))$, but $e_n(x)=f(b,c)$ so it is sufficient to check that $b=e_n(x,x)$ and $c= t \mapsto e_n(x, i_{n-1}(t))$, which are the coherence conditions of $X_n$.
Furthermore we need for $x'$ in $X_n$ corresponding to a tuple $(b',c'),f)$, $e_n(x,x') = f( e_n(x',x'),t \mapsto e_n(x',i_{n-1}(t)))$, which is true because by definition of $e_n$, $e_n(x,x') = f(b',c')$ and $b' = f'(b',c') =e_n(x',x')$ while $c'(t) = f'( e_{n-1}(t,t), s \mapsto e_{n-1}(s,t)) = e_n(x', i_{n-1}(t))$ by definition of $e_n$ and $i_{n-1}$.
The fact that $e_{n+1}(i_n(x_1),i_n(x_2))=e_n(x_1,x_2)$ follows from unwinding the definitions.
Indeed, let $x_1 = ((b_1,c_1),f_1)$ and let $f_2=((b_2,c_2),f_2))$ and then $e_{n+1} (i_n ( x_1),i_n(x_2)) = f_1 ( e_n (x_2,x_2), t \mapsto e_n(x_2, i_{n-1}(t)) =f_1(b_2,c_2)= e_n(x_1,x_2)$ by the coherence conditions for $b_2,c_2$ and the definition of $e_n$, and then the definition of $e_n$.
Consider the map $j_n: Y_{n+1} \to Y_{n}$ that sends $(b,c)$ to $( b, c \circ i_n)$. Consider also the map $k_n: Z_{n} \to Z_{n+1}$ by exponentiating $j_n$.
I claim that if $i_{n} ((b,c),f)= ((b',c'),f')$ then $j_n(b',c') = (b,c)$ and $k_n(f)=f'$.
The first statement is simply the fact that $ b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$, which follow from the definition of $e_n$ and the coherence conditions for $b$ and $c$ respectively.
The second statement follows immediately from the construction of $i_n$, which forces $f' ( \beta,\gamma) = $f( \beta , \gamma \circ i_n) = \alpha(x)$ which is exactly $f \circ k_n$.
To do this, it is sufficient to prove that the projection map $X_n \to Z_n$ is surjective. In other words, given $f: Y_n \to I$, construct $(b,c) \in Y_n$ satisfying the coherence conditions. Clearly we must have $c(x) = f( e_n(x,x), t \mapsto e_n(x,t))$. Then we can take $b$ to be any fixed point of $f(b,c)$.
What maps can we construct this way? We get all the maps defined by elements of $Z_n$ for all $n$. These, in turn, are all continuous maps that depend only on the data in $Y_n$. By the coherence conditions and the definition of $e$, the projection of $x \in X_n$ to $Y_n$ encode $e(x,x),t\mapsto e(x,t)$ for $t$ in $X_{n-1}$. Using the definition of $e$ and the compatibility of $e$ with $i_{n-1}$, one can see that it also encodes $t \mapsto e(t,x)$ for $t \in Y_{n-1}$. The compatibility of $i_n$ with $e$ and with $j_n$ ensure that $Y_n$ continues to encode this data for all $x$. Hence all continuous functions that depend only on $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$, which seem to me to be all the "obvious" ones that we construct, are in the image of the constructed map $X \to I^X$.
The significance of the spaces $Y_n$ and $Z_n$ is that $Z_n$ is the set of continuous $I$-valued functions of $x \in X$ that depend only on $ e(x,x), e(x,t)$$e(t,x), e(x,x), e(x,t)$ for $t \in X_n$. $Y_n$ is the space of possible values of the pairtriple $e(x,x), e(x,t)$$e(t,x), e(x,x), e(x,t)$, so that $Z_n$ is the space of functions on $Y_n$. Then $X_{n+1}$ is constructed so that it maps surjectively to $Z_n$, so all functions of $Z_n$ come from elements of $X_n$, and maps to $Y_n$, so that all functions of $Z_n$ can be extended to functions on $X_n$. For simplicity and canonicality, we define $X_{n+1}$ to be a subset of $Y_n \times Z_n$, defined by coherence condition to ensure the desired relationship between $Y_n$ and $e$. Then we can freely add also the functions $e(t,x)$ for $t\in X_n$ to $Z_n$, as they are already continuous functions of $e(x,x)$, $e(x,t)$ by construction of $X_n$.
Let $Y_n$ be the subset of $ I \times I^{X_n}$$I^{X_n} \times I \times I^{X_n}$ consisting of tuples (a,b,c) satisfying the coherence conditions:
$(a \circ i_{n-1} , b, c \circ i_{n-1}) \in Y_{n-1}$
For all $x$ in $X_{n}$ which corespond to a tuple $(a',b',c'),f'$ in $Y_{n-1} \times Z_{n-1}$, $a(x)= f'( a \circ i_{n-1} ,b, c \circ i_{n-1})$.
In the case $n=0$, so $X_n = \emptyset$, this is vacuous, and so we need not introduce $Y_{-1}$ and $Z_{-1}$.
Let $Z_n= I^{Y_n}$.
Let $X_{n+1} $ be the subset of $Y_n \times Z_n$ consisting of tuples $(b,c)\in Y_n, f\in Z_n$$(a,b,c)\in Y_n, f\in Z_n$ satisfying the following two coherence conditions:
$b= f(b,c)$$b= f(a,b,c)$
For all $x$ in $X_n$, $c(x) = f( e_n(x,x), t \mapsto e_n(x,t))$$c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$
Let $i_n: X_n \to X_{n+1}$ send $x= ((b',c'),f)$$x$ to $(e_n(x,x), t \mapsto e_n(x,t) ) , (\beta,\gamma)\mapsto f'(\beta, \gamma \circ i_{n-1} )$$( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t) ) , f(\alpha,\beta,\gamma)=\alpha(x)$.
Let $e_{n+1}: X_{n+1} \times X_{n+1} \to I$ send $((b_1,c_1),f_1),((b_2,c_2),f_2)$$((a_1,b_1,c_1),f_1),((a_2,b_2,c_2),f_2)$ to $f_1(b_2,c_2)$$f_1(a_2,b_2,c_2)$.
Let us first check that for $x$ in $X_n$ corresponding to a tuple $(b,c),f$$(a,b,c),f$ in $Y_{n-1} \times Z_{n-1}$, $i_n(x)$ satisfies the coherence conditions.
For $x$$x'$ in $X_n$ which corresponds to a tuple $(b,c),f$$(a',b',c'),f'$, we need $e_n(x,x) = f( e_n(x,x), t \mapsto e_n(x, i_{n-1}(t))$, but $e_n(x)=f(b,c)$ so$e_n(x',x) = f'(t \mapsto e_n( i_{n-1}(t),x), e_n(x,x), t \mapsto e_n(x,i_{n-1}(t))$. So it is sufficient to check that $a(t) = e_n( i_{n-1}(t),x)$, $b=e_n(x,x)$, and $c= t \mapsto e_n(x, i_{n-1}(t))$$c=e_n(x,i_{n-1}(t))$, which arefollow from the coherence conditionsdefinition of $X_n$.
Furthermore we need for $x'$ in $X_n$ corresponding to a tuple $(b',c'),f)$, $e_n(x,x') = f( e_n(x',x'),t \mapsto e_n(x',i_{n-1}(t)))$, which is true because by$e_n$ and respectively the definition of $e_n$$i_{n-1}$, the coherence condition for $e_n(x,x') = f(b',c')$$b$, and $b' = f'(b',c') =e_n(x',x')$ while $c'(t) = f'( e_{n-1}(t,t), s \mapsto e_{n-1}(s,t)) = e_n(x', i_{n-1}(t))$ bythe definition of $e_n$ and $i_{n-1}$ as well as the coherence condition for $c$.
The fact that $e_{n+1}(i_n(x_1),i_n(x_2))=e_n(x_1,x_2)$ follows from unwinding the definitions.
Indeed, let $x_1 = ((b_1,c_1),f_1)$ and let $f_2=((b_2,c_2),f_2))$ and then $e_{n+1} (i_n ( x_1),i_n(x_2)) = f_1 ( e_n (x_2,x_2), t \mapsto e_n(x_2, i_{n-1}(t)) =f_1(b_2,c_2)= e_n(x_1,x_2)$ by the coherence conditions for $b_2,c_2$ and the definition of $e_n$, and then the definition of $e_n$.
Consider the map $j_n: Y_{n+1} \to Y_{n}$ that sends $(b,c)$$(a,b,c)$ to $( b, c \circ i_n)$$(a \circ i_n, b, c \circ i_n)$. Consider also the map $k_n: Z_{n} \to Z_{n+1}$ by exponentiating $j_n$.
I claim that if $i_{n} ((b,c),f)= ((b',c'),f')$$i_{n+1} ((a,b,c),f)= ((a',b',c'),f')$ then $j_n(b',c') = (b,c)$$j_n(a',b',c') = (a,b,c)$ and $k_n(f)=f'$.
The first statement is simply the fact that $ b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$$ a(t)= t \mapsto e_n(i_{n-1}(t),x), b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$, the first of which followfollows from the definition of $e_n$ and the second and third from the definition of $e_n$ and the coherence conditions for $b$ and $c$ respectively.
The second statement follows immediately fromis simply the construction ofstatement that for $i_n$$ (\alpha,\beta,\gamma)$ in $Y_{n+1}$, $f( \alpha \circ i_n, \beta , \gamma \circ i_n) = \alpha(x)$ which forcesis the coherence condition for $f' ( \beta,\gamma) = $f( \beta$a$ applied to , \gamma \circ i_n) = \alpha(x)$ which is exactly $f \circ k_n$$\alpha$.
To do this, it is sufficient to prove that the projection map $X_n \to Z_n$ is surjective. In other words, given $f: Y_n \to I$, construct $(b,c) \in Y_n$$(a,b,c) \in Y_n$ satisfying the coherence conditions. Clearly we must have $c(x) = f( e_n(x,x), t \mapsto e_n(x,t))$$c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$. For now, let $b$ be a parameter. Define $c_m: X_m \to I$ for $m$ from $0$ to $n$ such that $c_n= c$, $c_m = c_{m+1} \circ i_m$, then define $a_m: X_m \to I$ so that $a_0$ is the empty function and $a_m(x)$ for $x \in X_m$ corresponding to $(a',b',c',f')$ is $f'(a_{m-1}, b, c_{m-1})$. Then by the definition of $i_{m-1}$, $a_{m-1} =a_m \circ i_{m-1}$. Now each $a_m$ is a continuous function of the previous, so $a_n$ is a continuous function of $b$, hence $b\mapsto f(a_n,b,c)$ has a fixed point. So we canmay take $a=a_n$, $b$ to be anythe fixed point ofand $f(b,c)$$c=c$.
What maps can we construct this way? We get all the maps defined by elements of $Z_n$ for all $n$. These, in turn, are all continuous maps that depend only on the data in $Y_n$. By the coherence conditions and the definition of $e$, the projection of $x \in X_n$ to $Y_n$ encode $e(x,x),t\mapsto e(x,t)$$t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$ for $t$ in $X_{n-1}$. Using the definition of $e$ and the compatibility of $e$ with $i_{n-1}$, one can see that it also encodes $t \mapsto e(t,x)$ for $t \in Y_{n-1}$. The compatibilityconstruction of $i_n$ with $e$ and withthe projection $j_n$$j_n: Y_{n+1} to Y_n$ ensure that $Y_n$it continues to encode this data for all $x$. Hence all continuous functions that depend only on $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$, which seem to me to be all the "obvious" ones that we construct, are in the image of the constructed map $X \to I^X$.
The significance of the spaces $Y_n$ and $Z_n$ is that $Z_n$ is the set of continuous $I$-valued functions of $x \in X$ that depend only on $ e(x,x), e(x,t)$ for $t \in X_n$. $Y_n$ is the space of possible values of the pair $e(x,x), e(x,t)$, so that $Z_n$ is the space of functions on $Y_n$. Then $X_{n+1}$ is constructed so that it maps surjectively to $Z_n$, so all functions of $Z_n$ come from elements of $X_n$, and maps to $Y_n$, so that all functions of $Z_n$ can be extended to functions on $X_n$. For simplicity and canonicality, we define $X_{n+1}$ to be a subset of $Y_n \times Z_n$, defined by coherence condition to ensure the desired relationship between $Y_n$ and $e$. Then we can freely add also the functions $e(t,x)$ for $t\in X_n$ to $Z_n$, as they are already continuous functions of $e(x,x)$, $e(x,t)$ by construction of $X_n$.
Let $Y_n$ be $ I \times I^{X_n}$. Let $Z_n= I^{Y_n}$.
Let $X_{n+1} $ be the subset of $Y_n \times Z_n$ consisting of tuples $(b,c)\in Y_n, f\in Z_n$ satisfying the following two coherence conditions:
$b= f(b,c)$
For all $x$ in $X_n$, $c(x) = f( e_n(x,x), t \mapsto e_n(x,t))$
Let $i_n: X_n \to X_{n+1}$ send $x= ((b',c'),f)$ to $(e_n(x,x), t \mapsto e_n(x,t) ) , (\beta,\gamma)\mapsto f'(\beta, \gamma \circ i_{n-1} )$.
Let $e_{n+1}: X_{n+1} \times X_{n+1} \to I$ send $((b_1,c_1),f_1),((b_2,c_2),f_2)$ to $f_1(b_2,c_2)$.
Let us first check that for $x$ in $X_n$ corresponding to a tuple $(b,c),f$ in $Y_{n-1} \times Z_{n-1}$, $i_n(x)$ satisfies the coherence conditions.
For $x$ in $X_n$ which corresponds to a tuple $(b,c),f$, we need $e_n(x,x) = f( e_n(x,x), t \mapsto e_n(x, i_{n-1}(t))$, but $e_n(x)=f(b,c)$ so it is sufficient to check that $b=e_n(x,x)$ and $c= t \mapsto e_n(x, i_{n-1}(t))$, which are the coherence conditions of $X_n$.
Furthermore we need for $x'$ in $X_n$ corresponding to a tuple $(b',c'),f)$, $e_n(x,x') = f( e_n(x',x'),t \mapsto e_n(x',i_{n-1}(t)))$, which is true because by definition of $e_n$, $e_n(x,x') = f(b',c')$ and $b' = f'(b',c') =e_n(x',x')$ while $c'(t) = f'( e_{n-1}(t,t), s \mapsto e_{n-1}(s,t)) = e_n(x', i_{n-1}(t))$ by definition of $e_n$ and $i_{n-1}$.
The fact that $e_{n+1}(i_n(x_1),i_n(x_2))=e_n(x_1,x_2)$ follows from unwinding the definitions.
Indeed, let $x_1 = ((b_1,c_1),f_1)$ and let $f_2=((b_2,c_2),f_2))$ and then $e_{n+1} (i_n ( x_1),i_n(x_2)) = f_1 ( e_n (x_2,x_2), t \mapsto e_n(x_2, i_{n-1}(t)) =f_1(b_2,c_2)= e_n(x_1,x_2)$ by the coherence conditions for $b_2,c_2$ and the definition of $e_n$, and then the definition of $e_n$.
Consider the map $j_n: Y_{n+1} \to Y_{n}$ that sends $(b,c)$ to $( b, c \circ i_n)$. Consider also the map $k_n: Z_{n} \to Z_{n+1}$ by exponentiating $j_n$.
I claim that if $i_{n} ((b,c),f)= ((b',c'),f')$ then $j_n(b',c') = (b,c)$ and $k_n(f)=f'$.
The first statement is simply the fact that $ b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$, which follow from the definition of $e_n$ and the coherence conditions for $b$ and $c$ respectively.
The second statement follows immediately from the construction of $i_n$, which forces $f' ( \beta,\gamma) = $f( \beta , \gamma \circ i_n) = \alpha(x)$ which is exactly $f \circ k_n$.
To do this, it is sufficient to prove that the projection map $X_n \to Z_n$ is surjective. In other words, given $f: Y_n \to I$, construct $(b,c) \in Y_n$ satisfying the coherence conditions. Clearly we must have $c(x) = f( e_n(x,x), t \mapsto e_n(x,t))$. Then we can take $b$ to be any fixed point of $f(b,c)$.
What maps can we construct this way? We get all the maps defined by elements of $Z_n$ for all $n$. These, in turn, are all continuous maps that depend only on the data in $Y_n$. By the coherence conditions and the definition of $e$, the projection of $x \in X_n$ to $Y_n$ encode $e(x,x),t\mapsto e(x,t)$ for $t$ in $X_{n-1}$. Using the definition of $e$ and the compatibility of $e$ with $i_{n-1}$, one can see that it also encodes $t \mapsto e(t,x)$ for $t \in Y_{n-1}$. The compatibility of $i_n$ with $e$ and with $j_n$ ensure that $Y_n$ continues to encode this data for all $x$. Hence all continuous functions that depend only on $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$, which seem to me to be all the "obvious" ones that we construct, are in the image of the constructed map $X \to I^X$.
The significance of the spaces $Y_n$ and $Z_n$ is that $Z_n$ is the set of continuous $I$-valued functions of $x \in X$ that depend only on $e(t,x), e(x,x), e(x,t)$ for $t \in X_n$. $Y_n$ is the space of possible values of the triple $e(t,x), e(x,x), e(x,t)$, so that $Z_n$ is the space of functions on $Y_n$. Then $X_{n+1}$ is constructed so that it maps surjectively to $Z_n$, so all functions of $Z_n$ come from elements of $X_n$, and maps to $Y_n$, so that all functions of $Z_n$ can be extended to functions on $X_n$. For simplicity and canonicality, we define $X_{n+1}$ to be a subset of $Y_n \times Z_n$, defined by coherence condition to ensure the desired relationship between $Y_n$ and $e$.
Let $Y_n$ be the subset of $I^{X_n} \times I \times I^{X_n}$ consisting of tuples (a,b,c) satisfying the coherence conditions:
$(a \circ i_{n-1} , b, c \circ i_{n-1}) \in Y_{n-1}$
For all $x$ in $X_{n}$ which corespond to a tuple $(a',b',c'),f'$ in $Y_{n-1} \times Z_{n-1}$, $a(x)= f'( a \circ i_{n-1} ,b, c \circ i_{n-1})$.
In the case $n=0$, so $X_n = \emptyset$, this is vacuous, and so we need not introduce $Y_{-1}$ and $Z_{-1}$.
Let $Z_n= I^{Y_n}$.
Let $X_{n+1} $ be the subset of $Y_n \times Z_n$ consisting of tuples $(a,b,c)\in Y_n, f\in Z_n$ satisfying the following two coherence conditions:
$b= f(a,b,c)$
For all $x$ in $X_n$, $c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$
Let $i_n: X_n \to X_{n+1}$ send $x$ to $( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t) ) , f(\alpha,\beta,\gamma)=\alpha(x)$.
Let $e_{n+1}: X_{n+1} \times X_{n+1} \to I$ send $((a_1,b_1,c_1),f_1),((a_2,b_2,c_2),f_2)$ to $f_1(a_2,b_2,c_2)$.
Let us first check that for $x$ in $X_n$ corresponding to a tuple $(a,b,c),f$ in $Y_{n-1} \times Z_{n-1}$, $i_n(x)$ satisfies the coherence conditions.
For $x'$ in $X_n$ which corresponds to a tuple $(a',b',c'),f'$, we need $e_n(x',x) = f'(t \mapsto e_n( i_{n-1}(t),x), e_n(x,x), t \mapsto e_n(x,i_{n-1}(t))$. So it is sufficient to check that $a(t) = e_n( i_{n-1}(t),x)$, $b=e_n(x,x)$, and $c=e_n(x,i_{n-1}(t))$, which follow from the definition of $e_n$ and respectively the definition of $i_{n-1}$, the coherence condition for $b$, and the definition of $i_{n-1}$ as well as the coherence condition for $c$.
The fact that $e_{n+1}(i_n(x_1),i_n(x_2))=e_n(x_1,x_2)$ follows from unwinding the definitions.
Consider the map $j_n: Y_{n+1} \to Y_{n}$ that sends $(a,b,c)$ to $(a \circ i_n, b, c \circ i_n)$. Consider also the map $k_n: Z_{n} \to Z_{n+1}$ by exponentiating $j_n$.
I claim that if $i_{n+1} ((a,b,c),f)= ((a',b',c'),f')$ then $j_n(a',b',c') = (a,b,c)$ and $k_n(f)=f'$.
The first statement is simply the fact that $ a(t)= t \mapsto e_n(i_{n-1}(t),x), b=e_n(x,x), c(t)= t \mapsto e_n(x,i_{n-1}(t))$, the first of which follows from the definition of $e_n$ and the second and third from the definition of $e_n$ and the coherence conditions for $b$ and $c$ respectively.
The second statement is simply the statement that for $ (\alpha,\beta,\gamma)$ in $Y_{n+1}$, $f( \alpha \circ i_n, \beta , \gamma \circ i_n) = \alpha(x)$ which is the coherence condition for $a$ applied to $\alpha$.
To do this, it is sufficient to prove that the projection map $X_n \to Z_n$ is surjective. In other words, given $f: Y_n \to I$, construct $(a,b,c) \in Y_n$ satisfying the coherence conditions. Clearly we must have $c(x) = f( t \mapsto e_n(t,x), e_n(x,x), t \mapsto e_n(x,t))$. For now, let $b$ be a parameter. Define $c_m: X_m \to I$ for $m$ from $0$ to $n$ such that $c_n= c$, $c_m = c_{m+1} \circ i_m$, then define $a_m: X_m \to I$ so that $a_0$ is the empty function and $a_m(x)$ for $x \in X_m$ corresponding to $(a',b',c',f')$ is $f'(a_{m-1}, b, c_{m-1})$. Then by the definition of $i_{m-1}$, $a_{m-1} =a_m \circ i_{m-1}$. Now each $a_m$ is a continuous function of the previous, so $a_n$ is a continuous function of $b$, hence $b\mapsto f(a_n,b,c)$ has a fixed point. So we may take $a=a_n$, $b$ to be the fixed point and $c=c$.
What maps can we construct this way? We get all the maps defined by elements of $Z_n$ for all $n$. These, in turn, are all continuous maps that depend only on the data in $Y_n$. By the coherence conditions and the definition of $e$, the projection of $x \in X_n$ to $Y_n$ encode $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$ for $t$ in $X_{n-1}$. The construction of the projection $j_n: Y_{n+1} to Y_n$ ensure that it continues to encode this data for all $x$. Hence all continuous functions that depend only on $t\mapsto e(t,x),e(x,x),t\mapsto e(x,t)$, which seem to me to be all the "obvious" ones that we construct, are in the image of the constructed map $X \to I^X$.
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