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Jan 16, 2021 at 7:20 history edited andpe CC BY-SA 4.0
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Jan 16, 2021 at 7:14 vote accept andpe
Jan 15, 2021 at 23:45 comment added Pietro Majer By definition a map $Y\to X^X$ is continuous iff for any $x\in X$ the composition with the evaluation at $x$ (projection on the $x$-coordinate of the product) is continuous. Thus the continuity of $X^X\ni f\mapsto f\circ g\in X^X$ is obvious, since $X^X\ni f\mapsto (f\circ g)(x)\in X$ is just $f\mapsto f(g(x))$, the evaluation at $g(x)$.
Jan 15, 2021 at 14:33 review Close votes
Jan 19, 2021 at 20:26
Jan 15, 2021 at 13:31 answer added Henno Brandsma timeline score: 3
Jan 15, 2021 at 7:00 comment added Gregory Arone If $A, B$ are sets, $X$ is a topological space, and $\alpha\colon A\to B$ is any function then the induced map $X^B\to X^A$, defined by $f\mapsto f\circ \alpha$ is continuous. This follows from the universal property of product. Taking $A=B=X$ you get the result that you want.
Jan 15, 2021 at 6:12 history asked andpe CC BY-SA 4.0