How can we know that two equations, e.g., $f(x) = x$ and $g(x) = 2x + 1$ describe the same geometrical object?

Question

The typical procedure to prove that these two can describe the same object seems to me to show that $f(x) + x + 1 = g(x)$. What isn't so clear to me, is why exactly does this constitute a proof that the graph of these two functions correspond to the same object? In other words, how come $h(f(x)) = g(x)$ guarantees that the equality in algebra also holds in geometry?

There seems to me to be a tremendous difference between algebra, where an equation constitutes equality, and geometry, where we must imagine we can superpose two objects in other to say they are equal. So, for instance, even if $h(f(x)) = g(x)$, wouldn't it be possible that, now no longer speaking of algebra but of geometry, a good enough zoom would show us that actually, h(f(x)) is actually slightly tilted in comparison to g(x)? Somehow, we know that because, $h(f(x)) = g(x)$, this isn't the case, but why?

Answer 1

How do you define "same object"? Those functions are different functions, they are not the same. In geometry we often like to say that two shapes are the "same object" or congruent if we can transform one into the other solely by (proper) rotations and translations. Sometimes we also allow reflections and stretches.

The function $f$ is composed of the set of pairs of points $\{(x,x)\}$ and the function $g$ is composed of the set of the pairs of points $\{(x,2x+1)\}.$ We can translate the points in $g:$ $$T_{y-1}(\{(x,2x+1)\}) = \{(x,2x)\}$$ Then we can rotate by the appropriate angle: $$R_\theta (T_{y-1}(\{(x,2x+1)\}) = R_\theta (\{x,2x\}) = \{(x,x)\}$$

So the sets are equivalent up to a translation and rotation. I'm skipping over details here, but this is the gist.

Answer 2

One interpretation of "same geometrical object" is "isometric".

Let $F$ be the graph of $f$ and $G$ be the graph of $g$.

Then $\phi: F \to G$ given by $\phi(x,x) = \left(\frac{x}{\sqrt5},2\frac{x}{\sqrt5}+1\right)$ is an isometry.

$F$ is clearly isometric to $\mathbb R$. But the natural parametrization of $G$ given by $t \mapsto (t,2t+1)$ is not an isometry because the derivative has norm $\sqrt 5$, hence the scaling above.