The nova-pset Package

Homework template for STEM coursework and problem sets. Spin-off of adaptable-pset with some modifications and additional functions.

Examples

Example front page

Example logo integrations

Logos

The example images use placeholder logos. To use your own logo, obtain it from your institution and pass it as an image:

#let logo = image("your-logo.png", height: 25pt)

Example Code

#import "@preview/nova-pset:0.1.0": *

#let class = "math 347h"
#let assignment = "Homework 4"
#let author = "Samyak Jain"
// To use a logo, add an image to this folder and replace none:
// #let logo = image("your-logo.png", height: 25pt)
#let logo = none
#let instructor = "Prof. Fernandough"
#let semester = "Fall 2025"
#let due-time = "September 25, 2025"

#show: homework.with(
  class: class,
  assignment: assignment,
  author: author,
  logo: logo,
  instructor: instructor,
  semester: semester,
  due-time: due-time,
  paper-size: "us-letter",
  accent-color: rgb("#1c2b39"),
)

#set enum(numbering: "a)")

#q(title: "Problem 1")[
    Prove that the composition of two surjective functions is surjective.
]

#b() Suppose that $f : A -> B$ and $g : B -> C$ are surjective functions. Then the composition $g compose f : A -> C$ is $g compose f (a) = g(f(a))$. We claim that $g(f(a))$ is surjective, or that

$ forall c in C, exists a in A "such that" g(f(a)) = c $

Let $c in C$ be arbitrary. Because $g$ is surjective, we know that

$ forall c in C, exists b in B "such that" g(b) = c $

So there exists a $b in B$ such that $g(b) = c$. Then, because $f$ is surjective, we know that

$ forall b in B, exists a in A "such that" f(a) = b $

So there exists an $a in A$ such that $f(a) = b$. Then $g(f(a)) = g(b) = c$. This means that the composition $g compose f$ is surjective.

#Q()