Onto function

Step 1: For finite sets of equal size, an onto function is also one-one. Step 2: So it becomes a permutation of (4) elements. Step 3: The number is (4!=24).

Step 1: In an onto function, every codomain element needs at least one preimage. Step 2: With (3) domain elements, it is impossible to cover (5) codomain elements. Step 3: If (|A|<|B|), the number of onto functions is (0).

Step 1: For any (c\in C), since (g) is onto, there is (b\in B) such that (g(b)=c). Step 2: Since (f) is onto, for that (b), there is (a\in A) such that (f(a)=b). Step 3: Then (g(f(a))=c), so the composition is onto.

Step 1: If (g\circ f) is onto, then for every (c\in C), there is (a\in A) such that (g(f(a))=c). Step 2: Here (f(a)) is an element of (B), so (g) produces (c). Step 3: Hence every element of (C) is in the range of (g), so (g) is onto.

Step 1: If (g\circ f) is onto, a sufficient part of (g)'s domain covers (C). Step 2: This does not prove that (f) covers the whole of (B). Step 3: In reverse conclusions about composition, accept only what is logically forced.

Step 1: This is a continuous odd-degree polynomial. Step 2: Its derivative (5x^4+3x^2+1>0), so it is strictly increasing and covers both infinite directions. Step 3: For increasing polynomials, identifying the range becomes easier.

Step 1: (x^5-x) is a continuous odd-degree polynomial. Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: Even with turning points, continuity and end behavior can cover all of (\mathbb{R}).

Step 1: (x^3) is continuously increasing on ([-1,1]). Step 2: At (x=-1), it gives (-1), and at (x=1), it gives (1), so all intermediate values are obtained. Step 3: For a continuous increasing function on a closed interval, the range comes from endpoint values.

Step 1: The codomain is ([0,1]), so only target values from (0) to (1) must be checked. Step 2: For every (y\in[0,1]), (x=\sqrt[3]{y}) also lies in ([-1,1]). Step 3: Onto is checked against the codomain, not against extra values outside it.

Step 1: (\cos x) decreases from (1) to (-1) on ([0,\pi]). Step 2: Since it is continuous, every value in ([-1,1]) is obtained. Step 3: For trigonometric functions, the chosen interval determines the range.

Step 1: (\sin x) goes from (0) to (1) on \(\left[0,\frac{\pi}{2}\right]\). Step 2: The codomain ([-1,1]) has negative values, which are not obtained. Step 3: The same trigonometric formula can give different answers on different intervals.

Step 1: On this interval, (\sin x) takes all values from (-1) to (1). Step 2: The codomain is also ([-1,1]). Step 3: On a continuous monotonic part, the trigonometric range is clear.

Step 1: When an integer is divided by (3), the remainder is (0), (1), or (2). Step 2: For (0), take (n=0); for (1), take (n=1); for (2), take (n=2). Step 3: In remainder functions, check whether every possible remainder is obtained.

Step 1: (n^2) is never negative. Step 2: The codomain (\mathbb{Z}) contains integers like (-1), which are not obtained. Step 3: With integer domains, quickly identify missing values of the square function.

Step 1: (n^2) gives only perfect squares such as (0,1,4,9). Step 2: The codomain contains non-negative integers like (2) and (3), which are not perfect squares. Step 3: Even with a non-negative codomain, not every value must be obtained.

Step 1: (\lceil x\rceil) gives an integer for every (x). Step 2: For any (k\in\mathbb{Z}), taking (x=k) gives (\lceil x\rceil=k). Step 3: When the codomain of the ceiling function is (\mathbb{Z}), every integer is obtained.

Step 1: (\lceil x\rceil) is always an integer. Step 2: The codomain (\mathbb{R}) contains values like (\frac{1}{2}), which are not obtained. Step 3: For floor or ceiling functions with codomain (\mathbb{R}), check onto carefully.

Step 1: The sum of distances from (-2) and (2) is at least (4). Step 2: Between (-2) and (2), the value is (4), and outside this interval larger values are obtained. Step 3: Interpreting modulus as distance helps find the range.

Step 1: The minimum value of this modulus sum is (4). Step 2: As (|x|) becomes large, the value can grow without bound, so all values above (4) are obtained. Step 3: If the actual range equals the codomain, the function is onto.

Step 1: Since (x^2+1\ge1), (\sqrt{x^2+1}\ge1). Step 2: The codomain (\mathbb{R}) contains (0) and negative values, which are not obtained. Step 3: For square-root functions, the minimum of the inside expression determines the range.

Step 1: For any (y\ge1), set (y=\sqrt{x^2+1}). Step 2: Then (x^2=y^2-1), so (x=\sqrt{y^2-1}) is real. Step 3: If you can construct (x) from the target value, onto is proved.

Step 1: (x^2-2x+1=(x-1)^2). Step 2: Since (x\ge1), (x-1\ge0), and for every (y\ge0), take (x=1+\sqrt{y}). Step 3: Choose the preimage while respecting the domain restriction.

Step 1: (f(x)=(x-1)^2). Step 2: For every (y\ge0), taking (x=1-\sqrt{y}) gives (x\le1) and (f(x)=y). Step 3: A different branch of a quadratic can also be onto for a suitable codomain.

Step 1: (x^3+e^x) is continuous. Step 2: As (x\to-\infty), the (x^3) term makes the value go to (-\infty), and as (x\to\infty), both terms make it go to (\infty). Step 3: When a continuous function goes unbounded in both directions, it is onto (\mathbb{R}).

Step 1: For (x\ge0), (x^2) gives all non-negative real values. Step 2: For (x<0), (-x^2) gives all negative real values, and (0) is obtained from the first part. Step 3: For a piecewise function, find the range of each part and then take their union.