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In linear Algebra there is the Rank-nullity-theorem which states that for a linear map $f:V \rightarrow W$ we have

$$\dim{\operatorname{im}\,f}+\dim{\operatorname{ker}\,f}=\dim{V},$$

which implies

$$\dim{\operatorname{im}\,f} \leq \dim{V}.$$

Does this hold true for general maps $g:V \rightarrow W$?

I was thinking that this is probably not the case as there should be some surjective mapping $\mathbb{R}$ to $\mathbb{R}^2$ which exploits the density of $\mathbb{R}$. Perhaps something along the lines of a space-filling curve? I know there is the notion of fractal dimensionality and I realized that this question probably hinges on the definition of dimensionality. I would appreciate someone clarification or some reference, which might clarify this.

I apologize if this is a duplicate, I couldn't find anything similiar by search.

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    What does this have to do with covering spaces? – 2017-02-20
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    sorry that was a misclick – 2017-02-20
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    For a general map $im(f)$ doesn't need to be a linear subspace, so talking about its dimension might result meaningless – 2017-02-20
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    Sorry, I thought you were asking about maps between infinite dimensional vector spaces, hence the deleted answer. – 2017-02-20
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    let's say $g:\mathbb{R}^n\to\mathbb{R}^m$ with $m>n$. Can there be a surjective $g$ ? – 2017-02-20
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    Space filling curves are a well known example. So yes. – 2017-02-20
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    Use `\to` instead of `\arrow` – 2017-02-20
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    It is well known that the cardinality of $\mathbb{R}^m$ and $\mathbb{R}^n$ are equal for all $m$ and $n$. This says by definition that we in fact have bijections between them. – 2017-02-20
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    thanks, this is what I am looking for! Are there any well-known proofs for that? – 2017-02-20
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    I guess the whole question boils down to how you define dimensionality. If we are talking about the cardinality of the sets $V, W$ then obviously $W$ can't have a bigger cardinality than $V$ or $g$ wouldn't be a mapping. – 2017-02-20

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So the answer is clearly: no, as we can construct a bijection $g:\mathbb{R}^n \to \mathbb{R}^m$ with $m>n$.