If $g \circ f$ is injective and $f$ is surjective, then $g$ is injective.
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If $g\circ f$ is injective and $f$ is surjective then $g$ is injective
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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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Rigorous proof that surjectivity implies injectivity for finite sets
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If $f \circ g$ is onto then $f$ is onto and if $f \circ g$ is one-to-one then $g$ is one-to-one
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Let $X$ and $Y$ be non-empty sets and let $f:X \rightarrow Y$. Prove that $ f^{ −1} :P(Y ) → P(X) $ is injective if, and only if, $f$ is surjective.
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Injective functions also surjective?
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$I$ and $J$ are coprime ideals iff $x \to (x + I, x + J)$ is surjective.
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$f(g(x))=g(f(x))$ implies $f(c)=g(c)$ for some $c$
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Proving that a certain continuous function is surjective.
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Let $g : A \to B$ and $f : B \to C$ be functions. Let $f \circ g$ be onto. Are both $f$ and $g$ necessarily onto?
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If $g \circ f$ is surjective, show that $f$ does not have to be surjective?
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Showing function is injective iff $f(M_1 \cap M_2) = f(M_1)\cap f(M_2)$
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Prove that $f$ is one-to-one iff $f \circ h = f \circ k$ implies $h = k$.
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Show that $f$ is either injective or a constant function.
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Demonstrate that if $f$ is surjective then $X = f(f^{-1}(X))$
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$F:G\to B$ is an isomorphism between groups implies $F^{-1}$ is an isomorphism
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Proof of properties of injective and surjective functions.