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For example, I take differentiability, analyticity, and algebraicity(of a function). All(more or less) imply continuity. So when we define a differentiable function on $\mathbb R^n$ or an analytic function on $\mathbb C^n$, or a regular map on an affine space, we do not explicitly require that the functions are continuous. It follows automatically from the stronger condition.

But, when I look at the definitions in books of a global structure using sheaf theory, for a global definition of a morphism, ie on a differentiable manifold or an analytic space, or an abstract algebraic variety, the definition of a morphism requires a priori that the map be continuous, and then one requires that there is additionally a morphism of sheaves of algebras(of the suitable type of structure sheaves, depending on the local model used).

Why is this so? Is it something done for fancy, or is there a real need for the extra continuity assumption? I mean could things go wrong if this assumption is dropped?

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Sheaves are defined on open sets. So in order to have a behaved push-forward (which is needed before the morphism of sheaves, you need open sets to pull back to open sets, that is, continuity. – Andrea Ferretti May 19 '10 at 14:01
up vote 3 down vote accepted

As Andrea hints, if you start with sheaves then you need continuity to even begin talking about morphisms of sheaves.

However, if you're interesting in just defining, say, a smooth map between manifolds then you can simply write "$f \colon M \to N$ is smooth if, whenever $c \colon \mathbb{R} \to M$ is a smooth curve then $f \circ c \colon \mathbb{R} \to N$ is smooth". No assumption about continuity is needed there.

Indeed, once one gets to more exotic spaces, continuity becomes a hassle and is best left to one side. For example, the evaluation map $E \times E^* \to \mathbb{R}$ is smooth for any locally convex topological vector space, $E$, but is only continuous for $E$ a normed vector space.

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Let $M,N$ two manifolds and $f : M \to N$ a (set-theoretic) map. Then there are (at least) two definitions for $f$ to be smooth:

(1) For every ball $B \subseteq N$ the preimage $f^{-1}(B)$ can be covered with balls $C \subseteq M$ such that the induced maps $C \to B$ are smooth.

(2) $f$ is continuous and for every ball $B \subseteq N$ and every ball $C \subseteq f^{-1}(B)$ the induced map $C \to B$ is smooth.

Remark that in (1) it follows automatically that $f$ is continuous. However, the second statement in (2) does not imply continuity because it is possible that $f^{-1}(B)$ contains no ball at all, or just not enough.

The same is true for other subsheaves of continuous functions mentioned in the question.

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The simple answer is because you are describing your "spaces" (manifolds, variety etc.) as locally ringed spaces, in particular objects in the category of topological spaces + more data. The arrows in the category of topological spaces are precisely continuous maps, so this is where continuity comes in. To be more specific, let $C$ be any category, and let $S:C \to Cat$ be any pseudo-functor (weak 2-functor). "Pretend" that $S$ is the assignment of each object $c \in C_0$ its category of "sheaves of algebraic objects", e.g. if $C$ is topological spaces you could let $S$ be $S:X \mapsto Sh_{rings}(X)$ which associates to a space $X$ the category of sheaves of local rings over $X$. For any such $S$, one can take its Grothendieck construction, which yields a category fibred over $C$, $\int_C{S} \to C$. The objects of $\int_C{S}$ are pairs $(c,s)$ with $c \in C_0$ and $s \in S(C)_0$ and the maps $(c,s) \to (d,t)$ are pairs $(f,g)$ such that $f:c \to d$ in $C$ and $g:f^*(t) \to s$ in $S(C)$, and the functor $\int_C{S} \to C$ sends $(c,s)$ to $c$ and $(f,g)$ to $f$. If $S$ and $C$ are taken to be $Sh_{rings}$ and $Top$, then you get exactly the category of locally ringed spaces, for example. By construction, the "underlying morphism" of a morphism $(c,s) \to (d,t)$ is a morphism $f:c \to d$. If $C$ is $Top$, then of course, this means it is a morphism in $Top$, hence continuous.

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$S$ should be contravariant. – David Carchedi May 21 '10 at 14:53

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