Q.Show that the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=\frac{x}{x^2+1}\ \forall x\in\mathbb{R}$ is not onto.

A function is into when its range does not fill the whole codomain: $f(A)\subset B$.

For $f : A \to B$, the range is $f(A) \subseteq B$. The function is onto (surjective) when $f(A) = B$. It is into when at least one element of $B$ is …

True - onto $\Leftrightarrow$ range $=Y$.

A function $f:X\to Y$ is onto (surjective) if every element of $Y$ is an image of …

$f$ is onto (surjective) precisely when the range $f(A)$ equals the whole codomain $B$.

A function $f:A\to B$ is onto if every element of $B$ is the image of at least one element of $A$; that is, the range equals the codomain: …

Onto $\Leftrightarrow f(A) = B$.

A function $f : A \to B$ is called onto (surjective) if every element of the codomain $B$ is the image of at least one element of $A$. Equivalently, the range (image) of $f$ equals the whole codomain:

$f(A) = B.$ …

By definition, $f:X\to Y$ is onto exactly when every element of $Y$ is the image of some element of $X$, i.e. range $=Y$.

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Onto requires every element of the codomain to have a pre-image; find an element of N that $f$ never reaches.

$f:N\to N$, $f(x)=2x$. The range of $f$ is the set of even natural numbers $\{2,4,6,\dots\}$.

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$f$ is onto when $f(A)=B$.

A function $f:A\to B$ is onto (surjective) if every element of $B$ is the image of some element of $A$, i.e. the range …

Every real y is hit: choosing x = y/2 gives f(x) = y, so f is onto.

f : R → R, f(x) = 2x. A function is onto (surjective) if every element of the codomain has a pre-image.

Step 1: Take any y ∈ R (codomain).

Step 2: Choose x = y/2, which is a real number, so x ∈ R (domain).

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