I'm curious about the identity map on the space of all smooth maps (between two locally convex topological vector spaces in the sense of convenient calculi) when we equip the space with different topologies.
$$ \iota : \{f \in X \rightarrow Y : f \text{ is smooth}\} \rightarrow C^\infty(X,Y). $$
The space $\{f \in X \rightarrow Y : f \text{ is smooth}\}$ is just a subset of $X \rightarrow Y$ which is just the same as the product space $\Pi_{x \in X} Y$, i.e. $X$ worth of copies of $Y$. The topology on $\{f \in X \rightarrow Y : f \text{ is smooth}\}$ is the product topology i.e. in terms of functions this is pointwise convergence on $X \rightarrow Y$.
The space $C^\infty(X, Y)$ as a set is the same as the previous one but the topology is the initial topology given by maps $f \rightarrow y^* \circ f \circ c$ for every $c \in C^\infty(\mathbb{R}, X), y^* \in Y^*$. The result $y^* \circ f \circ c$ is an element of $C^\infty(\mathbb{R}, \mathbb{R})$ which carries the topology of compact convergence (see point 5 on wiki).
Is $\iota$ a smooth map?
More explicitly, given a smooth curve $\gamma \in C^\infty(\mathbb R, \{f \in X \rightarrow Y : f \text{ is smooth}\})$, is $\iota \circ \gamma$ smooth as well?
I have tried to write out all the definitions but it didn't look that the statement is obviously true. Maybe it is false but this is not my area of expertise, so I'm missing a bag of common counter examples I could throw at it.
