Equivalent Paths Let $f:[a,b] \rightarrow \mathbb{R^n}$ and $g:[c,d] \rightarrow \mathbb{R}^n$ be equivalent paths in $\mathbb{R}^n$. Here, we are considering $f$ and $g$ to be equivalent if there exists a continuous and strictly monotonic bijection $\phi:[a,b] \rightarrow [c,d]$ such that 
$$
f(t) = (g \circ \phi)(t) \; \; \forall \; \; t \in [a,b]
$$
Apostol in Mathematical Analysis (2nd Ed) claims on p 136 that equivalent paths have the same graph, but I can't possibly see how this can be so. For example,
$$
\mathrm{graph}(g) = \{(u, g(u)) | u \in [c, d] \}  = \{ (\phi(t), g(\phi(t)) | t \in \phi^{-1}[c,d] \} = \{(\phi(t), f(t)) | t \in [a,b] \}
$$

which is almost the graph of $f$ but because of the first coordinate is not the graph of $f$. I can only believe that I'm misunderstaning what he means by "graph". If "graph" of $f$ is defined as the image
of $f$ then, yes, the statement would be true, but that's not how the graph of a map is defined. 
Any ideas about what's going on here?
 A: You are right, there is no way that $f$ and $g$ have the same graph under those conditions, mostly because it may happen (very often) that $[a,b] \neq [c,d]$, so there exists a point on the graph of $f$ with the first coordinate in $[a,b] \backslash [c,d]$ that clearly cannot appear in the graph of $g$ (and vice versa if $[a,b] \subset [c,d]$). 
P.S. Your definition of the graph is very standard and correct. Unless the author uses a different definition (which I doubt a lot, since I've never seen a different one), the problem lies elsewhere.
You may ask yourself the question though : what you understood (and asked here) is maybe not what the author meant. Perhaps you should read more on that or post more information, such as the definition of equivalent paths. Maybe the only place where you are wrong is the second sentence of your question, "It follows that there exists a continuous and strictly monotonic bijection $\phi$..." maybe this is not true.
I would really need to have the definition of equivalent paths in hand though to say more about this question.
Hope that helps,
EDIT : Hm. Maybe the author is "saying" graph, but is "meaning" image. If you replace the word "graph" with "image", everything works just fine. It is a common mistake (for human minds' sake) to use the term graph to mean image. I guess you should just be happy and get some sleep. =D
Also note that this notion of equivalent paths is often said to be a reparametrization of a curve in differential geometry. These are not exactly the same, but they're similar in some sense.
A: I have accepted Patrick's answer as his response was essentially correct; however the confusion was not the result of an unintentional slip but, rather, the author's very nonstandard definition of graph. I found that previously in the text, Apostol defined the "graph" of $f:[a,b] \rightarrow \mathbb{R}^n$ to be the set of points $\{f(t) \in \mathbb{R}^n | t \in [a,b]\}$, or, in other words, the image of $f$.Thus the statement made in the text regarding the "graph" of $f$ are correct when this non-standard definition of graph is taken into account.
