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Proof of (1):

(a). Suppose $X$ and $Y$ are $G$-spaces, the action of $G$ on $X$ is free, and $X\to X/G$ is a principal bundle, then the space of $G$-equivariant maps $$ F(X,Y)^G $$ is the same thing as the space of sections of the fibration $X\times_G Y \to X/G$.

(b). If $E\to B$ is a Hurewicz fibration, with $B$ connected and the fiber contractible, then the space of sections is again contractible. This can be seen as follows: the adjunction property shows that $$ \text{sec}(E \to B) \to F(B,E) \to F(B,B) $$ is a fibration (the displayed fiber is the space of sections---it's the fiber over the basepoint of $F(B,B)$ given by the identity map of $B$---and the other two spaces are function spaces). Since $F$ is contractible the map $E\to B$ is a homotopy equivalence and therefore so is $F(B,E) \to F(B,B)$. Hence $\text{sec}(E\to B)$ is contractible.

(c). It follows from (a) and (b) that the space of sections of $X\times_G Y \to X/G$ is contractible whenever $Y$ is.

(d). In your case, $X = P$ and $Y = EG$. Hence, $F(P,EG)^G$ is contractible.

Proof of (2):

I'll prove something slightly less, which is enough for what you wish to have:

I claim that if $X \to X/G$ and $Y \to Y/G$ are principal $G$-bundles then the map $$ F(X,Y)^G \to F(X/G,Y/G) $$ is a Hurewicz fibration. This is the same thing as the map $$ \text{sec}(X\times_G Y\to X/G) \to \text{sec}(X/G\times Y/G\to X/G) $$ induced by the fibration $X\times_G Y \to X/G \times Y/G$.

This map of section spaces is the map of fibers which is induced by a map of fibrations over a common base space: the first fibration is $F(X/G, X\times_G Y) \to F(X/G,X/G)$ and the second one is $F(X/G, X/G\times Y/G) \to F(X/G,X/G)$. Since the map $F(X/G, X\times_G Y) \to F(X/G, X/G\times Y/G)$ is a fibration, it is formal to show that the map of section spaces is too.

In your situation we take $X = P$ and $Y = EG$. Then we see that the map $$ F(P,EG)^G \to F(X,BG) $$ is a fibration. The fiber at the point of $F(X,BG)$ defined by the classifying map for $P \to X$ is then your gauge group $\cal G$.

Proof of (1):

(a). Suppose $X$ and $Y$ are $G$-spaces, the action of $G$ on $X$ is free, and $X\to X/G$ is a principal bundle, then the space of $G$-equivariant maps $$ F(X,Y)^G $$ is the same thing as the space of sections of the fibration $X\times_G Y \to X/G$.

(b). If $E\to B$ is a Hurewicz fibration, with $B$ connected and the fiber contractible, then the space of sections is again contractible. This can be seen as follows: the adjunction property shows that $$ \text{sec}(E \to B) \to F(B,E) \to F(B,B) $$ is a fibration (the displayed fiber is the space of sections---it's the fiber over the basepoint of $F(B,B)$ given by the identity map of $B$---and the other two spaces are function spaces). Since $F$ is contractible the map $E\to B$ is a homotopy equivalence and therefore so is $F(B,E) \to F(B,B)$. Hence $\text{sec}(E\to B)$ is contractible.

(c). It follows from (a) and (b) that the space of sections of $X\times_G Y \to X/G$ is contractible whenever $Y$ is.

(d). In your case, $X = P$ and $Y = EG$. Hence, $F(P,EG)^G$ is contractible.

Proof of (2):

I'll prove something slightly less, which is enough for what you wish to have:

I claim that if $X \to X/G$ and $Y \to Y/G$ are principal $G$-bundles then the map $$ F(X,Y)^G \to F(X/G,Y/G) $$ is a Hurewicz fibration. This is the same thing as the map $$ \text{sec}(X\times_G Y\to X/G) \to \text{sec}(X/G\times Y/G\to X/G) $$ induced by the fibration $X\times_G Y \to X/G \times Y/G$.

This map of section spaces is the map of fibers which is induced by a map of fibrations over a common base space: the first fibration is $F(X/G, X\times_G Y) \to F(X/G,X/G)$ and the second one is $F(X/G, X/G\times Y/G) \to F(X/G,X/G)$. Since the map $F(X/G, X\times_G Y) \to F(X/G, X/G\times Y/G)$ is a fibration, it is formal to show that the map of section spaces is too.

In your situation we take $X = P$ and $Y = EG$. Then we see that the map $$ F(P,EG)^G \to F(P/G,BG) $$ is a fibration. The fiber at the point of $F(P/G,BG)$ defined by the classifying map for $P \to P/G$ is then your gauge group $\cal G$.

Proof of (1):

(a). Suppose $X$ and $Y$ are $G$-spaces, the action of $G$ on $X$ is free, and $X\to X/G$ is a principal bundle, then the space of $G$-equivariant maps $$ F(X,Y)^G $$ is the same thing as the space of sections of the fibration $X\times_G Y \to X/G$.

(b). If $E\to B$ is a Hurewicz fibration, with $B$ connected and the fiber contractible, then the space of sections is again contractible. This can be seen as follows: the adjunction property shows that $$ \text{sec}(E \to B) \to F(B,E) \to F(B,B) $$ is a fibration (the displayed fiber is the space of sections---it's the fiber over the basepoint of $F(B,B)$ given by the identity map of $B$---and the other two spaces are function spaces). Since $F$ is contractible the map $E\to B$ is a homotopy equivalence and therefore so is $F(B,E) \to F(B,B)$. Hence $\text{sec}(E\to B)$ is contractible.

(c). It follows from (a) and (b) that the space of sections of $X\times_G Y \to X/G$ is contractible whenever $Y$ is.

(d). In your case, $X = P$ and $Y = EG$. Hence, $F(P,EG)^G$ is contractible.

Proof of (2):

I'll prove something slightly less, which is enough for what you wish to have:

I claim that if $X \to X/G$ and $Y \to Y/G$ are principal $G$-bundles then the map $$ F(X,Y)^G \to F(X/G,Y/G) $$ is a Hurewicz fibration. This is the same thing as the map $$ \text{sec}(X\times_G Y\to X/G) \to \text{sec}(X/G\times Y/G\to X/G) $$ induced by the fibration $X\times_G Y \to X/G \times Y/G$.

This map of section spaces is the map of fibers which is induced by a map of fibrations over a common base space: the first fibration is $F(X/G, X\times_G Y) \to F(X/G,X/G)$ and the second one is $F(X/G, X/G\times Y/G) \to F(X/G,X/G)$. Since the map $F(X/G, X\times_G Y) \to F(X/G, X/G\times Y/G)$ is a fibration, it is formal to show that the map of section spaces is too.

In your situation we take $X = P$ and $Y = EG$. Then we see that the map $$ F(P,EG)^G \to F(X,BG) $$ is a fibration. The fiber at the point of $F(X,BG)$ defined by $P \to P/G$ is then your gauge group $\cal G$.

Proof of (1):

(a). Suppose $X$ and $Y$ are $G$-spaces, the action of $G$ on $X$ is free, and $X\to X/G$ is a principal bundle, then the space of $G$-equivariant maps $$ F(X,Y)^G $$ is the same thing as the space of sections of the fibration $X\times_G Y \to X/G$.

(b). If $E\to B$ is a Hurewicz fibration, with $B$ connected and the fiber contractible, then the space of sections is again contractible. This can be seen as follows: the adjunction property shows that $$ \text{sec}(E \to B) \to F(B,E) \to F(B,B) $$ is a fibration (the displayed fiber is the space of sections---it's the fiber over the basepoint of $F(B,B)$ given by the identity map of $B$---and the other two spaces are function spaces). Since $F$ is contractible the map $E\to B$ is a homotopy equivalence and therefore so is $F(B,E) \to F(B,B)$. Hence $\text{sec}(E\to B)$ is contractible.

(c). It follows from (a) and (b) that the space of sections of $X\times_G Y \to X/G$ is contractible whenever $Y$ is.

(d). In your case, $X = P$ and $Y = EG$. Hence, $F(P,EG)^G$ is contractible.

Proof of (2):

I'll prove something slightly less, which is enough for what you wish to have:

I claim that if $X \to X/G$ and $Y \to Y/G$ are principal $G$-bundles then the map $$ F(X,Y)^G \to F(X/G,Y/G) $$ is a Hurewicz fibration. This is the same thing as the map $$ \text{sec}(X\times_G Y\to X/G) \to \text{sec}(X/G\times Y/G\to X/G) $$ induced by the fibration $X\times_G Y \to X/G \times Y/G$.

This map of section spaces is the map of fibers which is induced by a map of fibrations over a common base space: the first fibration is $F(X/G, X\times_G Y) \to F(X/G,X/G)$ and the second one is $F(X/G, X/G\times Y/G) \to F(X/G,X/G)$. Since the map $F(X/G, X\times_G Y) \to F(X/G, X/G\times Y/G)$ is a fibration, it is formal to show that the map of section spaces is too.

In your situation we take $X = P$ and $Y = EG$. Then we see that the map $$ F(P,EG)^G \to F(X,BG) $$ is a fibration. The fiber at the point of $F(X,BG)$ defined by the classifying map for $P \to X$ is then your gauge group $\cal G$.

Proof of (1):

(a). Suppose $X$ and $Y$ are $G$-spaces, the action of $G$ on $X$ is free, and $X\to X/G$ is a principal bundle, then the space of $G$-equivariant maps $$ F(X,Y)^G $$ is the same thing as the space of sections of the fibration $X\times_G Y \to X/G$.

(b). If $E\to B$ is a Hurewicz fibration, with $B$ connected and the fiber contractible, then the space of sections is again contractible. This can be seen as follows: the adjunction property shows that $$ \text{sec}(E \to B) \to F(B,E) \to F(B,B) $$ is a fibration (the displayed fiber is the space of sections---it's the fiber over the basepoint of $F(B,B)$ given by the identity map of $B$---and the other two spaces are function spaces). Since $F$ is contractible the map $E\to B$ is a homotopy equivalence and therefore so is $F(B,E) \to F(B,B)$. Hence $\text{sec}(E\to B)$ is contractible.

(c). It follows from (a) and (b) that the space of sections of $X\times_G Y \to X/G$ is contractible whenever $Y$ is.

(d). In your case, $X = P$ and $Y = EG$. Hence, $F(P,EG)^G$ is contractible.

Proof of (2):

I'll prove something slightly less, which is enough for what you wish to have:

I claim that if $X \to X/G$ and $Y \to Y/G$ are principal $G$-bundles then the map $$ F(X,Y)^G \to F(X/G,Y/G) $$ is a Hurewicz fibration. This is the same thing as the map $$ \text{sec}(X\times_G Y\to X/G) \to \text{sec}(X/G\times Y/G\to X/G) $$ induced by the fibration $X\times_G Y \to X/G \times Y/G$.

This map of section spaces is the map of fibers which is induced by a map of fibrations over a common base space: the first fibration is $F(X/G, X\times_G Y) \to F(X/G,X/G)$ and the second one is $F(X/G, X\times_{G\times G} Y) \to F(X/G,X/G)$. Since the map $F(X/G, X\times_G Y) \to F(X/G, X/G\times Y/G)$ is a fibration, it is formal to show that the map of section spaces is too.

In your situation we take $X = P$ and $Y = EG$. Then we see that the map $$ F(P,EG)^G \to F(X,BG) $$ is a fibration. The fiber at the point of $F(X,BG)$ defined by $P \to P/G$ is then your gauge group $\cal G$.

(Note: we didn't need to make any assumptions on $G$ in this argument).

Proof of (1):

(a). Suppose $X$ and $Y$ are $G$-spaces, the action of $G$ on $X$ is free, and $X\to X/G$ is a principal bundle, then the space of $G$-equivariant maps $$ F(X,Y)^G $$ is the same thing as the space of sections of the fibration $X\times_G Y \to X/G$.

(b). If $E\to B$ is a Hurewicz fibration, with $B$ connected and the fiber contractible, then the space of sections is again contractible. This can be seen as follows: the adjunction property shows that $$ \text{sec}(E \to B) \to F(B,E) \to F(B,B) $$ is a fibration (the displayed fiber is the space of sections---it's the fiber over the basepoint of $F(B,B)$ given by the identity map of $B$---and the other two spaces are function spaces). Since $F$ is contractible the map $E\to B$ is a homotopy equivalence and therefore so is $F(B,E) \to F(B,B)$. Hence $\text{sec}(E\to B)$ is contractible.

(c). It follows from (a) and (b) that the space of sections of $X\times_G Y \to X/G$ is contractible whenever $Y$ is.

(d). In your case, $X = P$ and $Y = EG$. Hence, $F(P,EG)^G$ is contractible.

Proof of (2):

I'll prove something slightly less, which is enough for what you wish to have:

I claim that if $X \to X/G$ and $Y \to Y/G$ are principal $G$-bundles then the map $$ F(X,Y)^G \to F(X/G,Y/G) $$ is a Hurewicz fibration. This is the same thing as the map $$ \text{sec}(X\times_G Y\to X/G) \to \text{sec}(X/G\times Y/G\to X/G) $$ induced by the fibration $X\times_G Y \to X/G \times Y/G$.

This map of section spaces is the map of fibers which is induced by a map of fibrations over a common base space: the first fibration is $F(X/G, X\times_G Y) \to F(X/G,X/G)$ and the second one is $F(X/G, X/G\times Y/G) \to F(X/G,X/G)$. Since the map $F(X/G, X\times_G Y) \to F(X/G, X/G\times Y/G)$ is a fibration, it is formal to show that the map of section spaces is too.

In your situation we take $X = P$ and $Y = EG$. Then we see that the map $$ F(P,EG)^G \to F(X,BG) $$ is a fibration. The fiber at the point of $F(X,BG)$ defined by $P \to P/G$ is then your gauge group $\cal G$.