If pigs can fly, then I am Napoleon.

One of the easiest things that you can do is argue on a false premise; if a hypothesis is false, then its conclusion can be whatever you wish. This explains my title.

For a more concrete example, consider the following hypothesis: "$\infty$ is a number." It is clearly false, but assume that you think that it is true. Then $1+\infty=\infty$, which implies that $1=0$ (after subtracting $\infty$ from both sides). Therefore, we have shown that $1=0$, which in turn implies that any two numbers $x$ and $y$ are the same (use the following map: $\left(y-x\right)\cdot1+x=x$, where it is assumed, without loss of generality, that $y\geqslant x$). In other words, if we allow $\infty$ to be a number, then only one number will exist, which is absurd.

Real debaters defend the veracity of their arguments with irrefutable truths of which, hopefully, the opposition is ignorant. Fake debaters futilely exchange unsubstantiated (at best) biases ad nauseam. When you argue, first make sure that your premises are actually true and not a whim.

Geometry Headaches: "Abstract Nonsense"

Geometry has recently been giving me a slight headache, so I will write about it like I usually do in this situation. I will first write about the abstract formulation of tangent vectors.

Let $M$ be a smooth manifold, and let $p\in M$. A derivation at $p$ is a function $v:C^\infty\left(M\right)\to\mathbb R$ such that for all $f,g\in C^\infty\left(M\right)$, it is the case that
$$v\left(fg\right)=f\left(p\right)vg+g\left(p\right)vf.$$
The tangent space to $M$ at $p$, denoted $T_pM$, is the set of all derivations at $p$. Strangely, an element of $T_pM$ is called a tangent vector at $p$.

Define the geometric tangent space to $\mathbb R^n$ at $a$, denoted $\mathbb R_a^n$, to be a sort of copy of $\mathbb R^n$ but with its origin at $a\in\mathbb R^n$. Formally,
$$\mathbb R^n_a=\left\{a\right\}\times\mathbb R^n.$$
This is so that $\mathbb R^n_a\cap\mathbb R^n_b=\varnothing$ whenever $a\neq b$. For an element $\left(a,v\right)\in\mathbb R^n_a$, write $v_a$ or $\left.v\right|_a$ instead, and call it a geometric tangent vector at $a$.

Let $f\in C^\infty\left(\mathbb R^n\right)$, and recall that the derivative of $f$ in the direction of some $v\in\mathbb R^n$ at some $a\in\mathbb R^n$, denoted $\left.D_v\right|_af$, satisfies
$$\left.D_v\right|_af=\left.\frac d{dt}\right|_{t=0}f\left(a+tv\right).$$
It goes without saying (from calculus) that $\left.D_v\right|_a$ is linear and is a derivation at $a$.

Surprisingly, $\mathbb R^n_a$ and $T_a\mathbb R^n$ are isomorphic (with the map $v_a\mapsto\left.D_v\right|_a$).

Let $M$ and $N$ be smooth manifolds, and let $F:M\to N$ be a smooth map. For each $p\in M$, define a map
$$dF_p:T_pM\to T_{F\left(p\right)}N$$
and call it the differential of $F$ at $p$. Let $v\in T_pM$, let $f\in C^\infty\left(N\right)$, and define
$$dF_p\left(v\right)\left(f\right)=v\left(f\circ F\right).$$
We are now almost ready to talk about differentiating maps between manifolds.