I am trying to understand the following point. In Mumford's Geometric Invariant Theory book (p.3), we have $f:T\rightarrow X$ a morphism of schemes. Then it is told that, as a scheme over $T$, $X\times T$ has $(f,1_T)$ as a canonical section. I don't understand this sentence. As a beginner in algebraic geometry, I think I may not have the same definition of a section.
Could anyone help me to clarify this point?
To spell out Evans Gambit's comment, let $\mathsf{C}$ be a category, and let $X$ and $Y$ be two objects in the category. A fiber product of $X$ and $Y,$ if it exists, is an object $Z$ in $\mathsf{C}$ equipped with two morphisms $\pi_X : Z\to X$ and $\pi_Y : Z\to Y$ satisfying the following universal property: given any object $W$ in $\mathsf{C}$ and morphisms $f : W\to X$ and $g : W\to Y,$ there exists a unique morphism $h : W\to Z$ such that $\pi_X\circ h = f$ and $\pi_Y\circ h = g.$ This is often represented via a commutative diagram:
Such a $Z$ is often denoted by $X\times Y$ (note that any fiber product of $X$ and $Y$ is unique up to unique isomorphism), and the $h$ above might be represented as $h = (f,g)$ using the notation which Mumford uses.
Put another way, there is a canonical bijection \begin{align*} \operatorname{Hom}_\mathsf{C}(W,X\times Y)&\leftrightarrow\operatorname{Hom}_\mathsf{C}(W,X)\times\operatorname{Hom}_\mathsf{C}(W,Y) \end{align*} which sends a morphism $h : W\to X\times Y$ to the pair $(\pi_X\circ h,\pi_Y\circ h),$ and a pair $(f : W\to X, g : W\to Y)$ of morphisms to the unique morphism $h = (f,g)$ given by the universal property of $Z.$
Now, let's look at all this in the context of what you're reading. You have two schemes, $X$ and $T.$ One of the first results you prove in algebraic geometry is that fiber products of schemes exist, so that there is a scheme $X\times T$ together with morphisms of schemes $\pi_X : X\times T\to X$ and $\pi_T : X\times T\to T$ satisfying the universal property of the fiber product.
In particular, since you have a morphism $f : T\to X,$ and there is always the identity morphism $1_T : T\to T,$ this means that there is a unique map of schemes $(f,1_T) : T\to X\times T$ such that $\pi_X\circ (f,1_T) = f$ and $\pi_T\circ(f,1_T) = 1_T$ guaranteed by the universal property of the fiber product.
Now, you may think of $X\times T$ as a scheme over $T$ via the morphism $\pi_T : X\times T\to T$ which comes packaged together with $X\times T.$ A section of this morphism would be a map $\sigma : T\to X\times T$ such that $\pi_T\circ\sigma = 1_T.$ Notice that we already saw that $(f,1_T) : T\to X\times T$ satisfies that property, so it is indeed a section of $\pi_T$!
As a remark, this is canonical insofar as $f : T\to X$ is fixed for you. Without a given map $f : T\to X$ for you to use here which is natural in whatever sense you may like, there may be no such section of $\pi_T$ at all. On the other hand, if you have multiple maps from $T$ to $X$ and no reason to consider any one of them as more important than the others, you will have many different sections of $\pi_T$ (one for each map $T\to X$), but there wouldn't be any good reason to call any particular one canonical.