We've proved that if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ then $f \circ g$ is continuous at c.
I was wondering that if $g$ is continuous at $c$ and $f \circ g$ is continuous at c. Must $f$ be continuous at $g(c)$? I can't find a counter example to it, so I'm assuming it's true but having no luck proving it
Let $f$ and $g$ be functions from $\mathbb{R}$ to $\mathbb{R}$. That is, $$f:\mathbb{R}\to\mathbb{R}$$ $$g:\mathbb{R}\to\mathbb{R}$$
If $g$ is continuous at $c\in\mathbb{R}$ and $f\circ g$ is continuous at $c$, then $f$ is continuous at $g(c)$. Not true. Let's see why. We first go through the logic, then we provide an example.
Assume the following:
(i) $g$ is continuous at $c$
(ii) $f\circ g$ is continuous at $c$
By (i), for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in\mathbb{R}$, if $|x-c|<\delta$, then $|g(x)-g(c)|<\epsilon$.
By (ii), for each $\lambda>0$, there exists $\eta>0$ such that for each $x\in\mathbb{R}$, if $|x-c|<\eta$, then $|(f\circ g)(x)-(f\circ g)(c)|<\lambda$.
We want to show that for each $\xi>0$, there exists $\sigma>0$ such that for each $x\in\mathbb{R}$, if $|x-g(c)|<\sigma$, then $|f(x)-f(g(c))|<\xi$.
Let $\xi>0$ be arbitrary. Now we want to show that there exists $\sigma>0$ such that for each $x\in\mathbb{R}$, if $|x-g(c)|<\sigma$, then $|f(x)-f(g(c))|<\xi$.
We know the following:
(I) there exists $\delta>0$ such that for each $x\in\mathbb{R}$, if $|x-c|<\delta$, then $|g(x)-g(c)|<\xi$.
(II) there exists $\eta>0$ such that for each $x\in\mathbb{R}$, if $|x-c|<\eta$, then $|(f\circ g)(x)-(f\circ g)(c)|<\xi$.
In other words:
(I) If $x$ is close enough to $c$ (how close? $\delta$ close), then $g(x)$ is as close as we want to $g(c)$.
(II) If $x$ is close enough to $c$ (how close? $\eta$ close), then $(f\circ g)(x)$ is as close as we want to $(f\circ g)(c)$.
We have a problem! We cannot say anything about how $f$ behaves when $x$ is really close to $g(c)$!
Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $$ f(x)=\begin{cases} 1&\text{if }x\text{ is rational}\\ 0&\text{if }x\text{ is irrational} \end{cases} $$ This function is called Dirichlet's function.
Let $g:\mathbb{R}\to\mathbb{R}$ be defined by $$ g(x)=\begin{cases} \frac{1}{q}&\text{if }x\text{ is rational, in which case }x=\frac{p}{q}\text{ in lowest terms and }q>0\\ 0&\text{if }x\text{ is irrational} \end{cases} $$ This function is called Thomae's function. It is continuous at every irrational number!
Notice for each $x\in\mathbb{R}$, $(f\circ g)(x)=1$. It is a constant function! Therefore it is continuous.
Let $c\in\mathbb{R}$ be any (I mean any) irrational number. Is $f$ continuous at $c$? No. As a matter of fact, $f$ is discontinuous everywhere! But $g$ and $(f\circ g)$ are continuous at $c$. Therefore the above statement has been proven to be not true.
I hope my long-winded answer not only answered your question, but also informed you of the logic behind continuity, and what it truly means.
Try these at $0$: $$ \eqalign{g(x) &= x^2\cr f(x) &= \cases{1 & if $x \ge 0$\cr 2 & if $x < 0$\cr}}$$
G emphasized text denotes "g is continuous at c"
F denotes "f is continuous at g(c)"
C denotes "f o g is continuous at c"
Sentential Calculus says:
(((G --> (F --> C)) --> ((G ^ F) --> C))) Through Importation
Which is the axiom of closest resemblance to your situation, but unfortunately it doesn't say much about :
(((G ^ F) --> C) --> (G --> (F --> C)))
As nothing can be concluded by the converse of this statement.
So in other words, nothing can be conclude from "if g is continuous at c and f∘g is continuous at c."
What you're asking for is the characteristic property of embeddings resp. quotient maps, that is:
A map has the characteristic property of embeddings if: $$f\text{ continuous}\iff \iota\circ f\text{ continuous}$$ Especially this requires the embedding to be continuous.
A map has the characteristic property of quotient maps if: $$g\text{ continuous}\iff g\circ\pi\text{ continuous}$$ Especially this requires the quotient map to be continuous. (That is the situation of your question.)
Besides, the embedded spaces resp. 'quotiented' spaces are precisely what we call subspaces resp. quotient spaces...
