For finite $S$, if $f:S\rightarrow S$ is surjective, then $f$ is injective.
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Surjectivity implies injectivity and conversely
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Injective functions also surjective?
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If $[G:K]$ is finite, then $[H:H \cap K] = [G:K]$ iff $G = HK$ (Hungerford Proposition 4.8, Proof)
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Let $S$ be a finite set. Show that any function $f: S \rightarrow \mathbb{R}^n$ is continuous on S. What if $S$ is countably infinite?
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Show that $f$ is either injective or a constant function.
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Subset of a finite set is finite
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Show that $[K:F]_s = [K:L]_s [L:F]_s$ and $[K:F]_i = [K:L]_i [L:F]_i$.
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Let $f$ be a surjective homomorphism. Prove that $\ker(f)$ is a maximal ideal
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If $f \circ g$ is surjective, then f is surjective
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If $g \circ f$ is surjective, show that $f$ does not have to be surjective?
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Proof of $ f:I_n \rightarrow I_m \Rightarrow n = m$
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$f$ is continuous and open implies $f$ injective
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prove that $f:X\rightarrow Y$ is surjective if and only if $f(f^{-1}(C))=C$
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$f_n \rightarrow f$ in measure and $g$ be measurable and finite a.e. Prove that $f_ng \rightarrow fg$ in measure
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Prove composition of two surjections is surjection
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What can you say about $f$ and $g$ in the case that $fg$ is 1) Injective, 2) Surjective - Cohn - Classic Algebra Page 15
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Showing a function $f$ cannot be surjective