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Stream: learning: questions

Topic: Conceptual Mathematics: Exercises Help

Fintan Halpenny (May 22 2021 at 07:52):

image-2e6194a8-a887-4db6-b219-b7e87bced3d1.jpg

Fintan Halpenny (May 22 2021 at 07:53):

Hey all, I'm working through Conceptual Mathematics (and loving it). I'm getting a bit stuck on some exercises so I thought it might be helpful to start a topic for help on them if others are also going through the book :)

To kick it off, I'm having trouble interpreting exercise 3 in session 15 (page 179 of the 2nd edition). I'm a bit confused about the setup and the definitions of "evaluation at 0" and "iteration".

Ralph Sarkis (May 22 2021 at 08:59):

The elements in the left hand side are functions $f: \mathbb{N} \rightarrow Y$ making the square below commute.
image.png

The elements in the right hand side are elements $y \in Y$ viewed as functions $y: \mathbf{1} \rightarrow Y$.

The goal of the exercise is to show that the elements in the LHS are in correspondence with the elements in the RHS. Starting with $f: \mathbb{N} \rightarrow Y$ in the LHS, there is a straightforward way to obtain an element of $Y$, apply $f$ to $0 \in \mathbb{N}$ to obtain $f(0) \in Y$, this is "evaluation at 0".

The other direction is less straightforward. First, we observe that any $f: \mathbb{N} \rightarrow Y$ in the LHS is completely determined by $f(0)$. Indeed, by the commutativity of the square above, you can derive the value of $f(1)$ by $f(1) = f(\sigma(0)) = \beta(f(0))$, the value of $f(2)$ by $f(2) = f(\sigma(1)) = \beta(f(1))$ and so on. More concisely, you can define $f(n) = \beta^n(f(0))$ where $\beta^n$ denotes the iteration of $\beta$ $n$ times, i.e.: $\beta^n = \beta \circ \stackrel{n}{\cdots} \circ \beta$. Now, if you are given $y \in Y$, you can set $f(0) = y$ and iterate $\beta$ to find all the other values of $f$. This yields the direction LHS -> RHS.

Finally, the exercise asks you to show the two operation we just described are inverses. Namely, if you start with $y \in Y$, you define $f$ as above and you evaluate it at 0, you should obtain $y$. Conversely, if you start with $g$ in the LHS and you evaluate at 0 and define $f$ as above setting $f(0) = g(0)$ you should conclude $f = g$.

Fintan Halpenny (May 30 2021 at 08:32):

Thanks for the help @Ralph Sarkis :blush: it's all coming together with your explanation and diving further into the book 🤿