# Exercise 7.80#

Letβs check that Example 7.79 makes sense. Fix any topological space \((π, \textbf{Op}_π)\) and any subset \(π β πβ\) of the interval domain. Define \(H_π(π) := \{ π : π β© R β π \enspace | \enspace \text{π is continuous}\}\).

Is \(π»_π\) a presheaf? If not, why not; if so, what are the restriction maps?

Is \(π»_π\) a sheaf? Why or why not?

## Alternative answer#

Regarding `1.`

, for \(H_X\) to be a presheaf we need it to preserve identities and composition. The restriction maps are given by restricting a continuous function \(f\) on \(U β© R\) to a smaller open subset \(V β© R\), which again is a continuous function. These restriction maps should preserve identities because there will be no change in the continuous functions when they are restricted from \(U\) to \(U\), i.e. from a domain of \(π β© R\) to \(π β© R\). They should preserve composition because restricting a function from \(U β© R\) to \(W β© R\) should be no different than restricting it from \(U β© R\) to \(V β© R\) to \(W β© R\) where \(W β V β U\).

Regarding `2.`

, for \(H_X\) to be a sheaf there must be a unique gluing \(f\) of the continuous functions \(f_i\) and \(f_j\) that agree on the overlap of their domains \(U_i\) and \(U_j\). Since the functions are continuous, agreement on their overlap means that at the boundary between their definitions they will be equal, and hence any gluing will be continuous as well. An additional restriction from \(U\) to \(U β© R\) before or after the gluing should not change the fact that there is a unique gluing.