Proving that $\tilde{\phi}$ is injective if and only if whenever $\phi(x_1) = \phi(x_2)$, we have $x_1 \sim x_2$.

This question builds off a question I asked here. I'll restate the background for the question. I am taking the below proposition for granted.

Let $X$ be a set with an equivalence relation $\sim$. Let $\phi: X \to Y$ be a map of sets such that $x_1 \sim x_2$ implies $\phi(x_1) = \phi (x_2)$, and $\pi: X \to X/\sim$ the canonical quotient map sending $x \mapsto \overline{x}$, where $\overline{x}$ is the coset containing $x$. Then there exists a uniquely defined map $\tilde{\phi}: X/\sim \to Y$ such that $\phi = \tilde{\phi} \circ \pi$.

The map is defined as follows. Given $y \in X/\sim$, the map $\pi$ is surjective, so there exists $x \in X$ such that $\pi(x) = y$. Then define $\tilde{\phi}(y) = \phi(x)$. I proved that this map is well-defined, unique, and works as intended.

The specific result I'm trying to prove is:

Prove that $\tilde{\phi}$ is injective if and only if whenever $\phi(x_1) = \phi(x_2)$, we have $x_1 \sim x_2$.

Here is my attempt.

($\Rightarrow$) Suppose that $\tilde{\phi}$ is injective. Given $x_1, x_2 \in X$, suppose that $\phi(x_1) = \phi(x_2)$. Then, by commutativity of the aforementioned triangle, we have \begin{align*} \phi(x_1) = \phi(x_2) & \iff (\tilde{\phi} \circ \phi)(x_1) = (\tilde{\phi} \circ \phi)(x_2) \\ & \iff \tilde{\phi}(\pi(x_1)) = \tilde{\phi}(\pi(x_2)) \\ & \iff \pi(x_1) = \pi(x_2) \\ & \iff \overline{x_1} = \overline{x_2} \\ & \iff x_1 \sim x_2. \end{align*} Conversely, suppose that whenever $\phi(x_1) = \phi(x_2)$, we have $x_1 \sim x_2$. We suppose that $\tilde{\phi}(y_1) = \tilde{\phi}(y_2)$ for $y_1, y_2 \in X / \sim$. As $\pi$ is onto $X / \sim$, there exist $x_1, x_2 \in X$ such that $\pi(x_1) = y_1$ and $\pi(x_2) = y_2$. So \begin{align*} \tilde{\phi}(y_1) = \tilde{\phi}(y_2) & \iff \tilde{\phi}(\pi(x_1)) = \tilde{\phi}(\pi(x_2)) \\ & \iff (\tilde{\phi} \circ \pi)(x_1) = (\tilde{\phi} \circ \pi)(x_2) \\ & \iff \phi(x_1) = \phi(x_2) \\ & \iff x_1 \sim x_2 \end{align*} If $x_1 \sim x_2$, then $\overline{x_1} = \pi(x_1) = \pi(x_2) = \overline{x_2}$. In particular, $y_1 = y_2$, so $\tilde{\phi}$ is injective.

How does this proof look? Have I made any leaps of logic?

Solution 1:

• I assume that the aforementioned triangle will actually be aforementioned (and contains the maps $\phi,\widetilde\phi$, and $\pi$ in the only sensible arrangement.)

• In the first line you wrote $\widetilde\phi\circ\phi$ but I think you meant $\widetilde\phi\circ\pi$ (based on the next line).

Otherwise this looks good!