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Let's say $(X,\sigma)$ and $(X,\tau)$ are topological spaces, and $f$ is a continuous function from the former to the latter. (That is, the inverse images of elements of $\tau$ are elements of $\sigma$.) How would I write this? Saying:

Let $f:X\to X$ be continuous

clearly doesn't work, since they are meant to have separate topologies. However, saying:

Let $f:(X,\sigma)\to(X,\tau)$ be continuous

doesn't seem right, as the domain of $f$ is $X$, not the ordered pair $(X,\tau)$, and similarly for the codomain. Should I just do away with function notation completely, and say this?:

Let $f$ be a continuous function from the topological space $(X,\sigma)$ to the topological space $(X,\tau)$.

The same problem arises for metric spaces.

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You can use the second form:

Let $f\colon (X,\sigma)\to(X,\tau)$ be continuous.

You're right that in the universe of sets, taken literally this notation implies that $f$ is one of the four functions from $\{\{X\}, \{X,\sigma\}\}$ to $\{\{X\}, \{X,\tau\}\}$ — hardly what you mean. The highlighted statement implicitly means that $f$ is a morphism in the category $\mathsf{Top}$ of topological spaces with morphisms all continuous functions. Anyone will understand you. Via the usual "forgetful functor" from $\mathsf{Top}$ to $\mathsf{Sets}$, the structure (the topologies) fall off, and $f$ is revealed to be just a function from $X$ to $X$.

Of course, you can use the third form too, but it's verbose.

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The second and the third are both fine, and I consider them synomyms. It's quite common to "abuse notation" and say that the domain of the function is the whole structure instead of just the underlying set, if you're interested in structure preserving maps, like here. If the topologies are clear from context, they're often omitted, but here they should be mentioned explicitly as they have the same underlying set.

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I would like to suggest an alternative:

Let $f:X \to X$ be $(\sigma,\tau)$-continuous.

This also works if topology is replaced by metrics or norms:

Let $f:X \to X$ be $(d_0,d_1)$-continuous.

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